RD Sharma Volume 2 - Differential Equations Solution For Class 12 Mathematics

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

So, in this question, the order of the differential equation is 3, and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is x and the term is multiplied by itself so the given equation is non-linear.

Question 2.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this question, the order of the differential equation is 2, and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable y and its derivatives are multiplied with a constant or independent variable only so this equation is linear differential equation.

Question 3.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this question, we first need to remove the term because this can be written as which means a negative power.

So, the above equation becomes as

So, in this, the order of the differential equation is 3, and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself so the given equation is non-linear.

Question 4.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

In this question we will be raising both the sides to power 6 so as to remove the fractional powers of derivatives of the dependent variable y

So, the equation becomes as

1+=

So, in this the order of the differential equation is 2 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by itself and many other are also, so the given equation is non-linear.

Question 5.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this question the order of the differential equation is 2 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by itself so the given equation is non-linear.

Question 6.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Squaring on both sides, we get

Cubing on both sides

So, in this question, the order of the differential equation is 2, and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself so the given equation is non-linear.

Question 7.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Since this question has fractional powers, we need to remove them.

So, squaring on both sides, we get

So, in this equation, the order of the differential equation is 4, and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself, also the degree of the equation is 2 which must be one for the equation to be linear so the given equation is non-linear.

Question 8.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Since this question has fractional powers, we need to remove them.

So, squaring on both sides, we get

So, in this equation, the order of the differential equation is 1 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable y and its derivatives are multiplied with a constant or independent variable only so this equation is linear differential equation.

Question 9.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Here in this question the dependent variable is x, and thus the order of the equation is 2, and the degree of the equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable x and its derivatives are multiplied with a constant or independent variable only so this equation is linear differential equation.

Question 10.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Here in this question the dependent variable is t, and thus the order of the equation is 2, and the degree of the equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable t and its derivative is multiplied together so this equation is non-linear differential equation.

Question 11.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this equation the order of the differential equation is 2 and the degree of the differential equation is 3.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable y and its derivative is multiplied together , also y is multiplied by itself so this equation is non-linear differential equation.

Question 12.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this equation the order of the differential equation is 3 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

Here dependent variable y’s derivative is multiplied with itself , so this equation is non-linear differential equation.

Question 13.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

(xy² +x)dx + (y–x²y)dy=0

Answer:

The above equation can be written as

x(+1)dx=y(–1)dy

–1)y=x(+1)

=x+x

So, from this equation it is clear that order of the differential equation is 1 and the degree of the differential equation is also 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by y and also y is multiplied by itself, so the given equation is non-linear.

Question 14.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

The above equation can be written as

Since the power of y can’t be rational so squaring on both sides

So, the order of the above differential equation 1 and the degree of the differential equation is 2

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 15.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Since the power of is not rational we need to make it rational therefore cubing on both sides, we get

So, the order of the above differential equation 2 and the degree of the differential equation is 3.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 16.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

The above equation can be written as

2 =–3

Since the equation has rational powers, we need to remove them so squaring both sides we get

So, the order of the above differential equation 2 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 17.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Since the above equation has rational powers, we need to remove them so squaring on both sides.

So, the order of the above differential equation 2 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 18.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

First of all, we will rearrange the above equation as follows

Since the above equation has rational powers we need to remove them so squaring on both sides.

So, the order of the above differential equation 1 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 19.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

where

Answer:

First of all, we will rearrange the above equation as follows

y–x= here we have substituted the value of p and taken out from the root

Since the above equation has rational powers we need to remove them so squaring on both sides.

So, the order of the above differential equation 1 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 20.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Conceptof the question

So, the equation becomes as follows

So, the order of the above differential equation 1 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term y is multiplied by itself, so the given equation is non-linear.

Question 21.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Conceptof the question

So, in this question, the x of sin(x) is replaced by which means that the power of is not defined as it approaches to infinity by the above formula.

So, the order of the above differential equation 2 and the degree of the differential equation is not defined.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by itself, so the given equation is non-linear.

Question 22.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

(y")² +(y')²+ sin y =0

Answer:

Here in question

Conceptof the question

So, in this question, the x of sin(x) is replaced by y which means that the power of y is not defined as it approaches infinity by the above formula

So, the order of the above differential equation 2 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y, and the term y is multiplied by itself, so the given equation is non-linear.

Question 23.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this question the order of the differential equation is 2 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y and the term is multiplied by itself so the given equation is non-linear.

Question 24.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Conceptof the question

So, in this question, the x of sin(x) is replaced by y which means that the power of y is not defined as it approaches infinity by the above formula

So, the order of the above differential equation 3 and the degree of the differential equation is 1.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question the dependent variable is y, and the term y is multiplied by itself, so the given equation is non-linear.

Question 25.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

Conceptof the question

For the degree to be defined of any differential equation the euqtion must be expressible in the form of a polynomial.

But, in this question the degree of the differential equation is not defined because the term on the right hand side is not expressible in the form of a polynomial.

Thus, the order of the above equation is 2 whereas the degree is not defined.

Since the degree of the equation is not defined the equation is non-linear.

Question 26.

Determine the order and degree of each of the following differential equations. State also whether they are linear or non-linear.

Answer:

So, in this question, the order of the differential equation is 1, and the degree of the differential equation is 3.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constants or independent variable, then the equation is linear.

So, in this question, the dependent variable is y and the term is multiplied by itself so the given equation is non-linear.

Exercise 22.10

Question 1.

Solve the following differential equations :

Answer:

Formula:-

(i) if a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Here,

This is a linear differential equation, comparing it with

P = 2, Q = e^3x

I.F = e^∫Pdx

= e^∫2dx

= e^2x

multiplying both the sides by I.F

Integrating it with respect to x,

ye^2x = ∫e^5x dx + c

Question 2.

Solve the following differential equations :

Answer:

Formula:-

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Given:-

This is a linear differential equation, comparing it with

P = 2,

I.F = e^∫Pdx

= e^∫2Pdx

= e^2x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

Question 3.

Solve the following differential equations :

Answer:

Formula:-

(i) if a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Here,

This is a linear differential equation, comparing it with

P = 2, Q = 6e^x

I.F = e^∫Pdx

= e^∫2Pdx

= e^2x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ ye^2x = ∫6e^x e^2x dx + c

⇒ ye^2x = ∫6e^3x dx + c

⇒ ye^2x = 2e^3x + c

⇒ y = 2e^3x + ce^–2x

Question 4.

Solve the following differential equations :

Answer:

Formula:-

(i) if a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

This is a linear differential equation, comparing it with

P = 1, Q = e^–2x

I.F = e^∫Pdx

= e^∫Pdx

= e^x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ y(e^x) = ∫e^–2x e^x dx + c

⇒ y(e^x) = ∫e^–x dx + c

⇒ y(e^x) = –e^–2x + c

⇒ y = –e^–2x + c e^–x

Question 5.

Solve the following differential equations :

Answer:

Formula:-

(i) if a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Given:-

This is a linear differential equation, comparing it with

Q = 1

I.F = e^∫Pdx

= e^–logx

= x^–1

multiplying both the sides by I.F

integrating it with respect to x,

Question 6.

Solve the following differential equations :

Answer:

Formula:-

(i) if a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

This is a linear differential equation, comparing it with

P = 2, Q = 4x

I.F = e^∫Pdx

= e^∫2dx

= e^2x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ y(e^2x) = ∫4x.e^2x dx + c

⇒ y(e^2x) = 4(x∫e^2x dx– ∫ ( ∫e^2x dx)dx) + c

using integration by part

Question 7.

Solve the following differential equations :

Answer:

Formula:-

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Given:-

This is a linear differential equation, comparing it with

, Q =

I.F = e^∫Pdx

= e^logx

= x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ yx = ∫e^x xdx + c

⇒ yx = x∫e^x dx– ∫ ( ∫e^x dx)dx) + c

using integration by part

yx = xe^x–∫e^x dx + c

⇒ yx = xe^x–e^x + c

⇒ yx = (x–1)e^x + c

Question 8.

Solve the following differential equations :

Answer:

(i) iIf a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

(iv)

Given:-

This is a linear differential equation, comparing it with

I.F = e^∫Pdx

= (x² + 1)²

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ y(x² + 1)² = –∫ dx + c

⇒ y(x² + 1)² = –x + c

Question 9.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

Given:-

This is a linear differential equation, comparing it with

Q = log x

I.F = e^∫Pdx

= e^log|x|

= x, x>0

The solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ yx = ∫logx.x.dx + c

Question 10.

Solve the following differential equations :

Answer:

Formula:-

(i)∫[f(x) + f’(x)]e^xdx = f(x)e^x + c

(ii) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(iii) ∫dx = x + c

Given:

This is a linear differential equation, comparing it with

I.F = e^∫Pdx

= e^–log|x|

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

using formula(v)

⇒ y = e^x + cx

Question 11.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Given:-

This is a linear differential equation, comparing it with

Q = x³

I.F = e^∫Pdx

= e^logx

= x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ yx = ∫ x³xdx + c

Question 12.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

given:

This is a linear differential equation, comparing it with

P = 1, Q = sinx

I.F = e^∫Pdx

= e^∫dx

= e^x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ y e^x = ∫sinx. e^x dx + c

Question 13.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Given:-

This is a linear differential equation, comparing it with

P = 1, Q = cosx

I.F = e^∫Pdx

= e^∫dx

= e^x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c₁

⇒ y e^x = ∫cosx. e^x dx + c₁

let I = ∫ e^x cosxdx

= cosx∫ e^xdx ∫(sinx∫e^xdx)dx + c₂

using integrating by part

I = e^x cosx + ∫sinxe^xdx + c

= e^x cosx [sinx∫e^xdx∫(cosx∫e^xdx)dx] + c₂

⇒ I = e^x cosx + sinxe^x–I + C₂

⇒ 2I = (cosx + sinx)e^x + C₂

putting I

Question 14.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫dx = x + c

Given:-

This is a linear differential equation, comparing it with

P = 2, Q = sinx

I.F = e^∫Pdx

= e^∫2dx

= e^2x

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ y e^2x = ∫sinx. e^2x dx + c

Question 15.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫tanxdx = log|secx| + c

Given:-

This is a linear differential equation, comparing it with

P = – tanx, Q = – 2 sinx

I.F = e^∫^Pdx

= e^∫–tanxdx

= e^–log|secx|

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ ycosx = –2sinxcosxdx + c₁

⇒ ycosx = –sin2xdx + c₁

Question 16.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

Given:-

This is a linear differential equation, comparing it with

I.F = e^∫Pdx

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

let tan^–1x = t

so, ye^t = –∫te^t dt + c

= t∫ e^t dt–∫( e^t dt)dt + c

using integration by parts

y e^t = te^t –e^t + c

⇒ y = (t–1)ce^–t

Question 17.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

Given:-

This is a linear differential equation, comparing it with

P = tanx, Q = cosx

I.F = e^∫Pdx

= e^∫tanxdx

= e^log|secx|

= secx

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ ysecx = –∫cosx.secxdx + c

⇒ y = xcosx + Ccosx

Question 18.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

given:-

This is a linear differential equation, comparing it with

P = cotx, Q = x² cotx + 2x

I.F = e^∫Pdx

= e^∫cotxdx

= e^log|sinx|

= sinx

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ ysinx = ∫(x² cosx + 2xsinx)dx + c

⇒ ysinx = ∫(x² cosxdx + ∫2xsinxdx + c

⇒ ysinx = x²sinx + c

Question 19.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

(ii) ∫tanxdx = log|secx| + c

(iv) ∫cosxdx = sinx + c

given:-

This is a linear differential equation, comparing it with

P = tanx, Q = x²cos²x

I.F = e^∫Pdx

= e^∫tanxdx

= e^log|secx|

= secx

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

⇒ ysecx = ∫(x² cos²x(secx)dx + c

⇒ ysinx = ∫(x² cosxdx + c

⇒ ysecx = x²∫ cosxdx–∫(2x cosxdx)dx + c

using integrating by parts

y(secx) = x²sinx–2∫x² sinxdx + c

⇒ y(secx) = x²sinx–2(x∫ sinxdx–∫ sinxdx)dx + c

⇒ y(secx) = x²sinx + 2xcosx–2sinx + c

⇒ y = x²sinxcosx–2xcos²x–2sinxcos²x–2sinxcosx + ccosx

Question 20.

Solve the following differential equations :

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

Given:-

This is a linear differential equation, comparing it with

I.F = e^∫Pdx

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

so,

yt = ∫t.dt + c

Question 21.

Solve the following differential equations :

xdy = (2y + 2x⁴ + x²)dx

Answer:

(i) If a differential equation is ,

then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e^∫Pdx

Given:-

This is a linear differential equation, comparing it with

Q = 2x³ + x

I.F = e^∫Pdx

= e^–2logx

Solution of the equation is given by

y(I.F) = ∫Q.(I.F)dx + c

Question 22.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

Hence, the solution of the differential equation is,

Let

[Differentiating both sides]

By substituting this in the above integral, we get

We know

Thus, the solution of the given differential equation is

Question 23.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = y^–2 and Q = y^–3

The integrating factor (I.F) of this differential equation is,

We have

Hence, the solution of the differential equation is,

Let

[Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ xt = –{t log t – t} + c

⇒ xt = –t log t + t + c

Thus, the solution of the given differential equation is

Question 24.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = 10y²

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = y² [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ xy² = 2y⁵ + c

∴ x = 2y³ + cy^–2

Thus, the solution of the given differential equation is x = 2y³ + cy^–2

Question 25.

Solve the following differential equations:

(x + tan y)dy = sin 2y dx

Answer:

Given (x + tan y)dy = sin 2y dx

This is a first order linear differential equation of the form

Here, P = –cosec 2y and

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

[∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let tan y = t

⇒ sec²y dy = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

[∵ t = tan y]

Thus, the solution of the given differential equation is

Question 26.

Solve the following differential equations:

dx + xdy = e^–ysec²ydy

Answer:

Given dx + xdy = e^–ysec²ydy

This is a first order linear differential equation of the form

Here, P = 1 and e^–ysec²y

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^y [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ xe^y = tan y + c

∴ x = (tan y + c)e^–y

Thus, the solution of the given differential equation is x = (tan y + c)e^–y

Question 27.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = –tan x and Q = –2 sin x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

[∵ e^{log x} = x]

∴ I.F = cos x

Hence, the solution of the differential equation is,

Let cos x = t

⇒ –sinxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ yt = t² + c

[∵ t = cos x]

∴ y = cos x + c sec x

Thus, the solution of the given differential equation is y = cos x + c sec x

Question 28.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = cos x and Q = sin x cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^{sin x}

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ ye^t = te^t – e^t + c

⇒ ye^t × e^–t = (te^t – e^t + c)e^–t

⇒ y = t – 1 + ce^–t

∴ y = sin x – 1 + ce^{–sin x} [∵ t = sin x]

Thus, the solution of the given differential equation is y = sin x – 1 + ce^{–sin x}

Question 29.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = x² + 2

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

[∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall and

∴ y = (x² + 1)(x + tan^–1x + c)

Thus, the solution of the given differential equation is y = (x² + 1)(x + tan^–1x + c)

Question 30.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = cot x and Q = 2 sin x cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

[∵ t = sin x]

Thus, the solution of the given differential equation is

Question 31.

Solve the following differential equations:

Answer:

Given

[∵ x² – 1 = (x + 1)(x – 1)]

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have and

[∵ m log a = log a^m]

[∵ log a + log b = log ab]

[∵ e^{log x} = x]

Hence, the solution of the differential equation is,

We can write (x – 1)² = (x + 1)² – 4x

Recall and

Thus, the solution of the given differential equation is

Question 32.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = cos x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x² [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ yx² = x²sin x – 2{–x cos x + sin x} + c

⇒ yx² = x²sin x + 2x cos x – 2 sin x + c

Thus, the solution of the given differential equation is

Question 33.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = –1 and Q = xe^x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^–x

Hence, the solution of the differential equation is,

Recall

Thus, the solution of the given differential equation is

Question 34.

Solve the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = 2 and Q = xe^4x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^2x

Hence, the solution of the differential equation is,

Recall

Thus, the solution of the given differential equation is

Question 35.

Solve the differential equation given that when x = 2, y = 1.

Answer:

Given and when x = 2, y = 1

This is a first order linear differential equation of the form

Here, and Q = 2y

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = y^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

We know

⇒ xy^–1 = 2y + c

⇒ xy^–1 × y = (2y + c)y

∴ x = (2y + c)y

However, when x = 2, we have y = 1.

⇒ 2 = (2 × 1 + c) × 1

⇒ 2 = 2 + c

∴ c = 2 – 2 = 0

By substituting the value of c in the equation for x, we get

x = (2y + 0)y

⇒ x = (2y)y

∴ x = 2y²

Thus, the solution of the given differential equation is x = 2y²

Question 36.

Find one-parameter families of solution curves of the following differential equations:

, m is a given real number

Answer:

, m is a given real number

Given

This is a first order linear differential equation of the form

Here, P = 3 and Q = e^mx

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^3x

Hence, the solution of the differential equation is,

Case (1): m + 3 = 0 or m = –3

When m + 3 = 0, we have e^{(m + 3)x} = e⁰ = 1

⇒ ye^3x = x + c

⇒ ye^3x × e^–3x = (x + c)e^–3x

∴ y = (x + c)e^–3x

Case (2): m + 3 ≠ 0 or m ≠ –3

When m + 3 ≠ 0, we have

Recall

Thus, the solution of the given differential equation is

Question 37.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = –1 and Q = cos 2x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^–x

Hence, the solution of the differential equation is,

Let

⇒ 5I = e^–x(2 sin 2x – cos 2x)

By substituting the value of I in the original integral, we get

Thus, the solution of the given differential equation is

Question 38.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let

By substituting this in the above integral, we get

We know

∴ y = –e^–x + cx

Thus, the solution of the given differential equation is y = –e^–x + cx

Question 39.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = x³

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

We know

Thus, the solution of the given differential equation is

Question 40.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

Let t = log x

[Differentiating both sides]

By substituting this in the above integral, we get

We have

⇒ I.F = e^{log t}

⇒ I.F = t [∵ e^{log x} = x]

∴ I.F = log x [∵ t = log x]

Hence, the solution of the differential equation is,

Let t = log x

[Differentiating both sides]

By substituting this in the above integral, we get

We know

[∵ t = log x]

Thus, the solution of the given differential equation is

Question 41.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = x² + 2

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

[∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall and

∴ y = (x² + 1)(x + tan^–1x + c)

Thus, the solution of the given differential equation is y = (x² + 1)(x + tan^–1x + c)

Question 42.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = cos x and Q = e^{sin x} cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^{sin x}

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

[∵ t = sin x]

Thus, the solution of the given differential equation is

Question 43.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = –1 and Q = y

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^–y

Hence, the solution of the differential equation is,

Recall

⇒ xe^–y = –ye^–y – e^–y + c

⇒ xe^–y = –e^–y(y + 1) + c

⇒ xe^–y × e^y = [–e^–y(y + 1) + c] × e^y

∴ x = –(y + 1) + ce^y

Thus, the solution of the given differential equation is x = –(y + 1) + ce^y

Question 44.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, P = sec²x and Q = tan x sec²x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^{tan x}

Hence, the solution of the differential equation is,

Let tan x = t

⇒ sec²xdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ ye^t = te^t – e^t + c

⇒ ye^t × e^–t = (te^t – e^t + c)e^–t

⇒ y = t – 1 + ce^–t

∴ y = tan x – 1 + ce^{–tan x} [∵ t = tan x]

Thus, the solution of the given differential equation is y = tan x – 1 + ce^{–tan x}

Question 45.

Find one-parameter families of solution curves of the following differential equations:

e^–ysec²ydy = dx + xdy

Answer:

e^–ysec²ydy = dx + xdy

Given e^–ysec²ydy = dx + xdy

This is a first order linear differential equation of the form

Here, P = 1 and e^–ysec²y

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^y [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ xe^y = tan y + c

∴ x = (tan y + c)e^–y

Thus, the solution of the given differential equation is x = (tan y + c)e^–y

Question 46.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

Let t = log x

[Differentiating both sides]

By substituting this in the above integral, we get

We have

⇒ I.F = e^{log t}

⇒ I.F = t [∵ e^{log x} = x]

∴ I.F = log x [∵ t = log x]

Hence, the solution of the differential equation is,

Let t = log x

[Differentiating both sides]

By substituting this in the above integral, we get

We know

⇒ yt = t² + c

[∵ t = log x]

Thus, the solution of the given differential equation is

Question 47.

Find one-parameter families of solution curves of the following differential equations:

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = x log x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x² [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

Thus, the solution of the given differential equation is

Question 48.

Solve each of the following initial value problems:

y’ + y = e^x,

Answer:

y’ + y = e^x,

Given y’ + y = e^x and

This is a first order linear differential equation of the form

Here, P = 1 and Q = e^x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^x

Hence, the solution of the differential equation is,

Recall

However, when x = 0, we have

∴ c = 0

By substituting the value of c in the equation for y, we get

Thus, the solution of the given initial value problem is

Question 49.

Solve each of the following initial value problems:

, y(1) = 0

Answer:

, y(1) = 0

Given and y(1) = 0

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

∴ y = –log x – 1 + cx

However, when x = 1, we have y = 0

⇒ 0 = –log 1 – 1 + c(1)

⇒ 0 = –0 – 1 + c

⇒ 0 = –1 + c

∴ c = 1

By substituting the value of c in the equation for y, we get

y = –log x – 1 + (1)x

⇒ y = –log x – 1 + x

∴ y = x – 1 – log x

Thus, the solution of the given initial value problem is y = x – 1 – log x

Question 50.

Solve each of the following initial value problems:

, y(0) = 0

Answer:

, y(0) = 0

Given and y(0) = 0

This is a first order linear differential equation of the form

Here, P = 2 and Q = e^–2xsin x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^2x

Hence, the solution of the differential equation is,

Recall

⇒ ye^2x = –cos x + c

⇒ ye^2x × e^–2x = (–cos x + c) × e^–2x

∴ y = (–cos x + c)e^–2x

However, when x = 0, we have y = 0

⇒ 0 = (–cos 0 + c)e⁰

⇒ 0 = (–1 + c) × 1

⇒ 0 = –1 + c

∴ c = 1

By substituting the value of c in the equation for y, we get

y = (–cos x + 1)e^–2x

∴ y = (1 – cos x)e^–2x

Thus, the solution of the given initial value problem is y = (1 – cos x)e^–2x

Question 51.

Solve each of the following initial value problems:

, y(1) = 0

Answer:

, y(1) = 0

Given and y(1) = 0

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let

By substituting this in the above integral, we get

We know

∴ y = –e^–x + cx

However, when x = 1, we have y = 0

⇒ 0 = –e^–1 + c(1)

⇒ 0 = –e^–1 + c

∴ c = e^–1

By substituting the value of c in the equation for y, we get

y = –e^–x + (e^–1)x

∴ y = xe^–1 – e^–x

Thus, the solution of the given initial value problem is y = xe^–1 – e^–x

Question 52.

Solve each of the following initial value problems:

, y(0) = 0

Answer:

, y(0) = 0

Given and y(0) = 0

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

Hence, the solution of the differential equation is,

We know

However, when x = 0, we have y = 0

⇒ 0 = (0 + c)e⁰

⇒ 0 = (c) × 1

∴ c = 0

By substituting the value of c in the equation for x, we get

Thus, the solution of the given initial value problem is

Question 53.

Solve each of the following initial value problems:

, y(0) = 1

Answer:

, y(0) = 1

Given and y(0) = 1

This is a first order linear differential equation of the form

Here, P = tan x and Q = 2x + x²tan x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sec x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ y sec x = x²sec x + c

∴ y = x² + c cos x

However, when x = 0, we have y = 1

⇒ 1 = 0² + c(cos 0)

⇒ 1 = 0 + c(1)

∴ c = 1

By substituting the value of c in the equation for y, we get

y = x² + (1)cos x

∴ y = x² + cos x

Thus, the solution of the given initial value problem is y = x² + cos x

Question 54.

Solve each of the following initial value problems:

, y(0) = 1

Answer:

, y(0) = 1

Given and y(0) = 1

This is a first order linear differential equation of the form

Here, P = tan x and Q = 2x + x²tan x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sec x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ y sec x = x²sec x + c

∴ y = x² + c cos x

However, when x = 0, we have y = 1

⇒ 1 = 0² + c(cos 0)

⇒ 1 = 0 + c(1)

∴ c = 1

By substituting the value of c in the equation for y, we get

y = x² + (1)cos x

∴ y = x² + cos x

Thus, the solution of the given initial value problem is y = x² + cos x

Question 55.

Solve each of the following initial value problems:

,

Answer:

Given and

This is a first order linear differential equation of the form

Here, P = cot x and Q = 2 cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ yt = t² + c

[∵ t = sin x]

However, when, we have y = 0

⇒ 0 = 1 + c

∴ c = –1

By substituting the value of c in the equation for y, we get

[∵ sin²θ + cos²θ = 1]

∴ y = –cos x cot x

Thus, the solution of the given initial value problem is y = –cosec x cot x

Question 56.

Solve each of the following initial value problems:

,

Answer:

Given and

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ xy = x sin x + c

However, when, we have y = 1

∴ c = 0

By substituting the value of c in the equation for y, we get

∴ y = sin x

Thus, the solution of the given initial value problem is y = sin x

Question 57.

Solve each of the following initial value problems:

,

Answer:

Given and

This is a first order linear differential equation of the form

Here, P = cot x and Q = 4x cosec x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ y sin x = 2x² + c

∴ y = (2x² + c) cosec x

However, when, we have y = 0

By substituting the value of c in the equation for y, we get

Thus, the solution of the given initial value problem is

Question 58.

Solve each of the following initial value problems:

, y = 0 when

Answer:

xi. , y = 0 when

Given and

This is a first order linear differential equation of the form

Here, P = 2 tan x and Q = sin x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = sec²x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ ysec²x = sec x + c

∴ y = cos x + c cos²x

However, when, we have y = 0

∴ c = –2

By substituting the value of c in the equation for y, we get

y = cos x + (–2)cos²x

∴ y = cos x – 2cos²x

Thus, the solution of the given initial value problem is y = cos x – 2cos²x

Question 59.

Solve each of the following initial value problems:

, y = 2 when

Answer:

, y = 2 when

Given and

This is a first order linear differential equation of the form

Here, P = –3 cot x and Q = sin 2x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = cosec³x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ ycosec³x = 2(–cosec x) + c

⇒ ycosec³x = –2cosec x + c

∴ y = –2sin²x + csin³x

However, when, we have y = 2

⇒ 2 = –2(1)² + c(1)³

⇒ 2 = –2 + c

∴ c = 4

By substituting the value of c in the equation for y, we get

y = –2sin²x + (4)sin³x

∴ y = –2sin²x + 4sin³x

Thus, the solution of the given initial value problem is y = –2sin²x + 4sin³x

Question 60.

Solve each of the following initial value problems:

,

Answer:

Given and

This is a first order linear differential equation of the form

Here, P = cot x and Q = 2 cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ yt = t² + c

[∵ t = sin x]

However, when, we have y = 0

⇒ 0 = 1 + c

∴ c = –1

By substituting the value of c in the equation for y, we get

[∵ sin²θ + cos²θ = 1]

∴ y = –cos x cot x

Thus, the solution of the given initial value problem is y = –cosec x cot x

Question 61.

Solve each of the following initial value problems:

dy = cos x (2 – y cosec x)dx

Answer:

dy = cos x (2 – y cosec x)dx

Given dy = cos x (2 – y cosec x)dx

This is a first order linear differential equation of the form

Here, P = cot x and Q = 2 cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Let sin x = t

⇒ cosxdx = dt [Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ yt = t² + c

[∵ t = sin x]

Thus, the solution of the given differential equation is

Question 62.

Solve each of the following initial value problems:

, tan x ≠ 0 given that y = 0 when

Answer:

, tan x ≠ 0 given that y = 0 when

Given and

This is a first order linear differential equation of the form

Here, P = cot x and Q = 2x + x² cot x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ y sin x = x²sin x + c

However, when, we have y = 0

By substituting the value of c in the equation for y, we get

Thus, the solution of the given initial value problem is

Question 63.

Find the general solution of the differential equation.

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x² [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

Thus, the solution of the given differential equation is

Question 64.

Find the general solution of the differential equation

Answer:

Given

This is a first order linear differential equation of the form

Here, P = –1 and Q = cos x

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = e^–x

Hence, the solution of the differential equation is,

Let

⇒ I = e^–x(sin x – cos x) – I

⇒ 2I = e^–x(sin x – cos x)

By substituting the value of I in the original integral, we get

Thus, the solution of the given differential equation is

Question 65.

Solve the differential equation

Answer:

Given

This is a first order linear differential equation of the form

Here, and Q = 3x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = x^–1 [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

We know

⇒ yx^–1 = 3x + c

⇒ yx^–1 × x = (3x + c)x

∴ y = (3x + c)x

Thus, the solution of the given differential equation is y = (3x + c)x

Question 66.

Find the particular solution of the differential equation y ≠ 0 given that x = 0 when

Answer:

Given

This is a first order linear differential equation of the form

Here, P = cot y and Q = 2y + y²cot y

The integrating factor (I.F) of this differential equation is,

We have

∴ I.F = sin y [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ x sin y = y² sin y + c

∴ x = y² + c cosec y

However, when, we have x = 0.

By substituting the value of c in the equation for x, we get

Thus, the solution of the given differential equation is

Question 67.

Solve the following differential equation

Answer:

Given

This is a first order linear differential equation of the form

Here, and

The integrating factor (I.F) of this differential equation is,

We have

Hence, the solution of the differential equation is,

Let cot^–1y = t

[Differentiating both sides]

By substituting this in the above integral, we get

Recall

⇒ xe^t = –{te^t – e^t} + c

⇒ xe^t = –te^t + e^t + c

⇒ xe^t × e^–t = (–te^t + e^t + c)e^–t

⇒ x = –t + 1 + ce^–t

[∵ t = cot^–1y]

Thus, the solution of the given differential equation is

Exercise 22.11

Question 1.

The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Answer:

Let the surface area of the balloon be S.

∴ S = 4πr²

According to the question,

⇒

⇒ 8πr

⇒ 8πrdr = ktdt

Integrating both sides, we have

⇒ 8π∫rdr = k∫tdt

⇒

⇒ ……(1)

Given, we have r = 1 unit when the t = 0 sec

Putting the value in equation (1)

∴

⇒ 4π (1)² = k × 0 + c

⇒ c = 4π ……(2)

Putting the value of c in equation (1) we have,

……(3)

Given, we have r = 2 units when t = 3 sec

∴

⇒

⇒ ……(4)

Now, putting the value of k in equation (2),

We have,

⇒

∴

Question 2.

A population grows at the rate of 5% per year. How long does it take for the population to double?

Answer:

Let the initial population be P_o.

And the population after time t be P.

According to question,

⇒

Integrating both sides, we have

⇒ 20∫ = ∫dt

⇒ 20log|P| = t + c ……(1)

Given, we have P = P_o when t = 0 sec

Putting the value in equation (1)

∴ 20log|P| = t + c

⇒ 20log|P_o| = 0 + c

⇒ c = 20log|P_o| ……(2)

Putting the value of c in equation (1) we have,

20log|P| = t + 20log|P_o|

⇒ 20log|P| – 20log|P_o| = t

⇒ 20(log |P| – log|P_o|) = t

[]

⇒ 20log ( = t ……(3)

Now, for the population to be doubled

Let P = 2P_o at time t1

∴ t = 20log (

⇒ t1 = 20log (

⇒ t1 = 20 log2

∴ time required for the population to be doubled = 20 log2 years

Question 3.

The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?

[Given log_e 5 = 1.609, log_e 2 = 0.6931]

Answer:

Let the initial population be P_o .

And the population after time t be P.

According to question,

⇒

Integrating both sides, we have

⇒ ∫ = k∫dt

⇒ log|P| = kt + c ……(1)

Given, we have P = P_o when t = 0 sec

Putting the value in equation (1)

∴ log|P| = kt + c

⇒ log|P_o| = 0 + c

⇒ c = log|P_o| ……(2)

Putting the value of c in equation (1) we have,

log|P| = kt + log|P_o|

⇒ log|P| – log|P_o| = k t

⇒ (log |P| – log|P_o|) = kt []

⇒ log ( = kt ……(3)

Now, the population doubled in 25 years.

Let P = 2P_o at t = 25 years

∴ kt = log (

⇒ k×25 = log (

⇒ k = ……(4)

Now, equation (3) becomes,

Now, let t1 be the time for the population to increase from 100000 to 500000

⇒

⇒ (log5 = 1.609 and log2 = 0.6931)

⇒ t1 = 58

∴ The time require for the population to be 500000 = 58 years.

Question 4.

In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 20000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let the count of bacteria be C at any time t.

According to question,

⇒ where k is a constant

⇒

Integrating both sides, we have

⇒ ∫ = k∫dt

⇒ log|C| = kt + a ……(1)

Given, we have C = 100000 when t = 0 sec

Putting the value in equation (1)

∴ log|C| = kt + a

⇒ log|100000| = 0 + a

⇒ a = log|100000| ……(2)

Putting the value of a in equation (1) we have,

log|C| = kt + log|100000|

⇒ log|C| – log|100000| = k t []

⇒ log ( = kt ……(3)

Also, at t = 2 years, = 110000

From equation(3),we have

∴ kt = log (

⇒ k×2 = log (

⇒ k = ……(4)

Now, equation (3) becomes,

Now, let t1 be the time for the population to reach 200000

⇒

∴ The time require for the population to be 200000 = hours

Question 5.

If the interest is compounded continuously at 6% per annum, how much worth ₹ 1000 will be after ten years? How long will it take to double ₹ 1000?

[Given e^0.6 = 1.822]

Answer:

Let the principal, rate and time be Rs P, r and t years.

Also, let the initial principal be P_o.

⇒

Integrating both sides, we have

⇒ ∫∫dt

⇒ log|P| = t + c ……(1)

Now, at t = 0, P = P_o

log| P_o | = 0 + c

⇒ c = log| P_o |……(2)

Putting the value of c in equation (1) we have,

log|P| = t + log|P_o|

⇒ log|P| – log|P_o| = t

⇒ (log |P| – log|P_o|) = t []

⇒ log ( = t ……(3)

Now, P_o = 1000, t = 10years, r = 6

∴ log ( = ×10

⇒ log ( = 0.6

⇒

⇒ P = ×1000

⇒ P = 1.822×1000 (Given: = 1.822)

⇒ P = 1822

Rs 1000 will be Rs 1822 after 10 years at 6% rate.

Question 6.

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

[Given log_e 3 = 1.0986, e^2.1972 = 9]

Answer:

Let the count of bacteria be C at any time t.

According to question,

⇒ where k is a constant

⇒

Integrating both sides, we have

⇒ ∫ = k∫dt

⇒ log|C| = kt + a……(1)

Given, we have C = C₀ when t = 0 sec

Putting the value in equation (1)

∴ log|C| = kt + a

⇒ log| C₀| = 0 + a

⇒ a = log| C₀| ……(2)

Putting the value of a in equation (1) we have,

log|C| = kt + log|100000|

⇒ log|C| – log| C₀| = k t []

⇒ log ( = kt ……(3)

Also, at t = 5 years, C = 3C₀

From equation(3),we have

∴ kt = log (

⇒ k×5 = log (

⇒ k = ……(4)

Now, equation (3) becomes,

Now, let C1 be the number of bacteria present in 10 hours, as

⇒

Let the time be t1 for bacteria to be 10 times

⇒

∴ The time required for = hours

Question 7.

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 25000 in 2000, what will be the population in 2010?

Answer:

Let the initial population be P_o .

And the population after time t be P.

According to question,

⇒

Integrating both sides, we have

⇒ ∫ = k∫dt

⇒ log|P| = kt + logc……(1)

Given, we have P = 200000 when t = 1990

Putting the value in equation (1)

∴ log|200000| = k× 1990 + logc……(2)

we have P = 250000 when t = 2000

Putting the value in equation (1)

∴ log|250000| = k× 2000 + logc……(3)

On subtracting equation(2) from(3) we have,

log|250000| – log|200000| = k× (2000 – 1990)

⇒

⇒ ……(4)

Substituting the value of k from (4) in (2), we have

log|200000| = × 1990 + logc……(5)

Substituting the value of k, log c and t = 2010 in (2), we have

Question 8.

If the marginal cost of manufacturing a certain item is given by Find the total cost function C(x), given that C(0) = 100.

Answer:

⇒ dC = (2 + 0.15x)dx

Integrating both sides we have

⇒ ∫ dC = ∫ (2 + 0.15x)dx

⇒ ∫dC = 2∫dx + 0.15∫xdx

⇒ ……(1)

Now, given C = 100 when x = 0

⇒ 100 = 0 + 0 + k

⇒ k = 100……(2)

Putting the value of k in equation (1)

Question 9.

A ban pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.

[Take e^0.08≈1.0833]

Answer:

Let the principal, rate and time be Rs P, r and t years.

Also, let the initial principal be P_o.

⇒

Integrating both sides, we have

⇒ ∫∫dt

⇒ log|P| = t + c……(1)

Now, at t = 0, P = P_o

log| P_o | = 0 + c

⇒ c = log| P_o |……(2)

Putting the value of c in equation (1) we have,

log|P| = t + log|P_o|

⇒ log|P| – log|P_o| = t

⇒ (log |P| – log|P_o|) = t []

⇒ log ( = t ……(3)

Now, t = 1 year, r = 8%

∴ log ( = ×1

⇒ log ( = 0.08

⇒

(Given: = 1.0833)

⇒

∴ Percentage increase = 0.0833×100 = 8.33%

Question 10.

In a simple circuit of resistance R, self inductance L and voltage E, the current i at any time t is given by If E is constant and initially no current passes through the circuit, prove that

Answer:

We know that in a circuit of R, L and E we have,

⇒

We can see that it is a linear differential equation of the form

Where P = and Q =

I.F = e^∫Pdt

= e^dt

Solution of the given equation is given by

i × I.F = ∫Q × I.F dt + c

⇒ i × = ∫ × dt + c

⇒ i = + c ……(1)

Initially, there was no current

So, at i = 0, t = 0

Now, putting the value of c in equation (1)

i = –

i = (1 – )

Question 11.

The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Answer:

Let the quantity of mass at any time t be A.

According to the question,

⇒ where k is a constant

⇒

Integrating both sides, we have

⇒ ∫ = – k∫dt

⇒ log|A| = – kt + c……(1)

Given, the Initial quantity of masss be A₀ when the t = 0 sec

Putting the value in equation (1)

∴ log|A| = – kt + c

⇒ log| A₀| = 0 + c

⇒ c = log| A₀| ……(2)

Putting the value of c in equation (1) we have,

log|A| = – kt + log| A₀|

⇒ log|A| – log| A₀| = – k t []

⇒ log ( = – kt ……(3)

Let the mass becomes half at time t1, A =

From equation(3),we have

∴ – kt = log (

⇒ – k×t1 = log (

⇒ – k×t1 =

⇒ – k×t1 = – log 2

⇒ t1 =

∴ Required time = where k is the constant of proportionality.

Question 12.

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half – life is 1590 years. What percentage will disappear in one year?

Answer:

Let the quantity of radium at any time t be A.

According to question,

⇒ where k is a constant

⇒

Integrating both sides, we have

⇒ ∫ = – k∫dt

⇒ log|A| = – kt + c……(1)

Given, Initial quantity of radium be A₀ when t = 0 sec

Putting the value in equation (1)

∴ log|A| = – kt + c

⇒ log| A₀| = 0 + c

⇒ c = log| A₀| ……(2)

Putting the value of c in equation (1) we have,

log|A| = – kt + log| A₀|

⇒ log|A| – log| A₀| = – k t []

⇒ log ( = – kt ……(3)

Given its half life = 1590 years,

From equation(3),we have

∴ – kt = log (

⇒ – k×1590 = log (

⇒ – k×1590 =

⇒ – k×1590 = – log 2

⇒ k =

∴ The equation becomes

log ( = – t

log ( = – 0.9996 t

Percentage Disappeared = (1 – 0.9996) × 100 = 0.04 %

Question 13.

The slope of the tangent at a point P(x, y) on a curve is If the curve passes through the point (3, – 4), find the equation of the curve.

Answer:

Given the slope of the tangent =

We know that slope of tangent =

∴

⇒

Integrating both sides,

⇒

⇒ ……(1)

Now, the curve passes through (3, – 4)

So, it must satisfy the above equation

∴

⇒ ( – 4)² + (3)² = c1

⇒ 16 + 9 = c1

⇒ c1 = 25

Putting the value of c1 in equation (1)

∴

Question 14.

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation

Answer:

Given the differential equation

⇒

Integrating both sides we have,

⇒

⇒ log|y| + log|1 – y| = log|1 + x| + logc

⇒ log|y(1 – y)| = log|c(1 + x)|

⇒ y(1 – y) = c(1 + x) ……(1)

Since, the equation passes through (2,2), So,

2(1 – 2) = c(1 + 2)

⇒ – 2 = c×3

⇒ c =

Therefore, equation (1) becomes

y(1 – y) = (1 + x)

Question 15.

Find the equation of the curve passing through the point and tangent at any point of which makes an angle with the x – axis.

Answer:

The equation is

It is passing through

∴

⇒ 1 = 0 + c

⇒ c = 1

Putting the value of c in the above equation

∴

Question 16.

Find the curve for which the intercept cut – off by a tangent on the x – axis is equal to four times the ordinate of the point of contact.

Answer:

Let P(x,y) be the point of contact of tangent and curve y = f(x).

It cuts the axes at A and B so, the equation of the tangent at P(x,y)

Y – y = (X – x)

Putting X = 0

Y – y = (0 – x)

⇒ Y = y – x

So, A(0, y – x)

Now, putting Y = 0

0 – y = (X – x)

⇒ X = x – y

So, B(x – y,0)

Given, intercept on x – axis = 4× ordinate

⇒ x – y = 4y

⇒ y + 4y = x

⇒ + 4 =

⇒ = – 4

We can see that it is a linear differential equation.

Comparing it with

P = , Q = – 4

I.F = e^∫Pdy

= e^dy

= e ^{– logy}

Solution of the given equation is given by

x × I.F = ∫Q × I.F dy + logc

⇒ x × () = ∫ – 4 × dy + logc

⇒ = – 4 log y + log c

⇒ = log y ^{– 4} + logc

⇒ = log c y ^{– 4}

⇒ = c y ^{– 4}

Question 17.

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2(x + 1) = 2e^2x.

Answer:

Given slope at any point = y + 2x

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – 1, Q = 2x

I.F = e^∫Pdx

= e^{– dx}

= e ^{– x}

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × e ^{– x} = ∫ 2x × e ^{– x} dx + c

⇒ ye ^{– x} = 2∫ x × e ^{– x} dx + c

⇒ ye ^{– x} = – 2x e ^{– x –} 2 e ^{– x} + c

⇒y = – 2x – 2 + ce^x ……(1)

As the equation passing through origin,

0 = 0 – 2 + c× 1

⇒ c = 2

Putting the value of c in equation (1)

∴ y = – 2x – 2 + 2e^x

Question 18.

The tangent at any point (x, y) of a curve makes an angle with the x - axis. Find the equation of the curve if it passes through (1, 2).

Answer:

Question 19.

Find the equation of the curve such that the portion of the x - axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Answer:

Let P(x,y) be the point of contact of tangent and curve y = f(x).

It cuts the axes at A and B so, equation of tangent at P(x,y)

Y – y = (X – x)

Putting X = 0

Y – y = (0 – x)

⇒ Y = y – x

So, A(0, y – x)

Now, putting Y = 0

0 – y = (X – x)

⇒ X = x – y

So, B(x – y,0)

Given, intercept on x - axis = 4× ordinate

⇒ x – y = 2x

⇒ – y = x

⇒

⇒ – logx = logy + c ……(1)

As it passes through (1,2)

So, the point must satisfy the equation above

– log1 = log2 + c

⇒ 0 = log2 + c

⇒ c = – log2

Putting the value of c in equation (1)

– logx = logy – log2

⇒ log2 = logx + logy

⇒ log2 = logxy

⇒ xy = 2

Question 20.

Find the equation to the curve satisfying and passing through (1, 0).

Answer:

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – , Q = 1

I.F = e^∫Pdx

= e^dx

= e^{log|x + 1| – log|x|}

= e^log

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × = ∫ 1 × dx + c

⇒ y × = ∫ dx + c

⇒ y × = x + logx + c ……(1)

As the equation passing through (1,0),

0 = 1 + log1 + c

⇒ c = – 1

Putting the value of c in equation (1)

∴ y × = x + logx – 1

Question 21.

Find the equation of the curve which passes through the point (3, – 4) and has the slope at any point (x, y) on it.

Answer:

Given the slope of the tangent =

We know that slope of tangent =

∴

⇒

Integrating both sides,

⇒

⇒……(1)

Now, the curve passes through (3, – 4)

So, it must satisfy the above equation

∴

⇒ – 4 = (3)²× c

⇒ – 4 = 9c

⇒ c =

Putting the value of c in equation (1)

∴

Question 22.

Find the equation of the curve which passes through the origin and has the slope x + 3y – 1 at any point (x, y) on it.

Answer:

Given Slope of the equation at any point (x,y) = x + 3y – 1

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – 3, Q = x – 1

I.F = e^∫Pdx

= e^{– 3dx}

= e ^{– 3x}

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × e ^{– 3x} = ∫ (x – 1) × e ^{– 3x}dx + c

⇒ y × e ^{– 3x} = (x – 1) × e ^{– 3x} – ∫(1)e ^{– 3x} dx + c

⇒ y × e ^{– 3x} = (x – 1) × e ^{– 3x +} (e ^{– 3x} ) ⁺ c

⇒

⇒ ……(1)

As the equation passing through origin(0,0)

0 = 0 + + c

⇒ c =

Putting the value of c in equation (1)

∴

Question 23.

At every point on a curve, the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.

Answer:

Given the slope at any time = x + xy

We know that slope of tangent =

∴

⇒

Integrating both sides,

⇒

⇒ ……(1)

Now, the curve passes through (0,1)

So, it must satisfy the above equation

∴

⇒ log 2 = 0 + c

⇒ c = log2

Putting the value of c in equation (1)

∴

Question 24.

A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is and hence find the curve.

Answer:

⇒

It is a homogenous equation,

Putting y = kx

So,

Putting the value of k

Differentiating with respect to x,

⇒

Let (h,k) be the point where tangent passes through origin and the length is equal to h. So, equation of tangent at (h,k) is

⇒

⇒ 2ky – 2k² = cx – ch – 2hx + 2h²

⇒ x(c – 2h) – 2ky + 2k² –hc + 2h² = 0

⇒ x(c – 2h) – 2ky + 2(k² –2h) – hc = 0

⇒ x(c – 2h) – 2ky + 2(ch) – hc = 0 ( h² + k² = ch as (h,k) on th curve)

⇒ x(c – 2h) – 2ky + hc = 0

Now, Length of perpendicular as tangent from origin is

Hence, x² + y² = cx is the required curve.

Question 25.

Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with the x - axis is twice the abscissa of the point of contact.

Answer:

Let P(x,y) be the point of contact of tangent and curve y = f(x).

It cuts the axes at A and B so, the equation of the tangent at P(x,y)

Y – y = (X – x)

Now, putting Y = 0

0 – y = (X – x)

⇒ X = x – y

So, B(x – y,0)

Given, the distance between the foot of ordinate of the point of contact and the point of intersection of tangent and x - axis = 2x

BC = 2x

⇒ y = 2x

⇒

Integrating both sides we have

⇒logx = 2logy + c……(1)

As it passes through (1,2)

So, the point must satisfy the equation above

log1 = 2log2 + c

⇒ 0 = 2log2 + c

⇒ c = – 2log2

Putting the value of c in equation (1)

logx = 2logy – 2log2

⇒ logx = 2(logy – log2)

⇒

Question 26.

The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.

Answer:

Given the equation of normal at point (x,y) on the curve

∴

Now, the curve passes through (3,0)

Integrating both sides

It also passes through (3,4),so

Putting the value of c in equation(1)

Question 27.

The rate of increase of bacteria in a culture is proportional to the number of bacteria present, and it is found that the number doubles in 6 hours. Prove that the bacteria becomes eight times at the end of 18 hours.

Answer:

Let the count of bacteria be C at any time t.

According to the question,

⇒ where k is a constant

⇒

Integrating both sides, we have

⇒ ∫ = k∫dt

⇒ log|C| = kt + a……(1)

Given, we have C = C₀ when t = 0 sec

Putting the value in equation (1)

∴ log|C| = kt + a

⇒ log| C₀| = 0 + a

⇒ a = log| C₀| ……(2)

Putting the value of a in equation (1) we have,

log|C| = kt + log| C₀|

⇒ log|C| – log| C₀| = k t []

⇒ log ( = kt ……(3)

Also, at t = 6 years, C = 2C₀

From equation(3),we have

∴ kt = log (

⇒ k×6 = log (

⇒ k = ……(4)

Now, equation (3) becomes,

Now, C = 8C₀

⇒

⇒ t = 18

∴ The time required = 18 hours

Question 28.

Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one - half of the original amount of radium to decompose?

[Given log_e0.989 = 0.01106 and log_e2 = 0.6931]

Answer:

Let the quantity of radium at any time t be A.

According to the question,

⇒ where k is a constant

⇒

Integrating both sides, we have

⇒ ∫ = – k∫dt

⇒ log|A| = – kt + c……(1)

Given, Initial quantity of radium be A₀ when t = 0 sec

Putting the value in equation (1)

∴ log|A| = – kt + c

⇒ log| A₀| = 0 + c

⇒ c = log| A₀| ……(2)

Putting the value of c in equation (1) we have,

log|A| = – kt + log| A₀|

⇒ log|A| – log| A₀| = – k t []

⇒ log ( = – kt ……(3)

Given that the radium decomposes 1.1% in 25 years,

A = (100 – 1.1)% = 98.9% = 0.989 A₀ at t = 25 years

From equation(3),we have

∴ – kt = log (

⇒ – k×25 = log (

⇒ k = –

∴ The equation becomes

log ( = – t

Now,

∴ log ( = – t

⇒ log ( = – t

⇒ = – t

⇒ (log 2 = 0.6931 and log 0.989 = 0.01106)

⇒

⇒ t = 1567 years

Question 29.

Show that all curves for which the slope at any point (x, y) on it is are rectangular hyperbola.

Answer:

Given the slope of the tangent =

∴

It is a homogenous equation,

Putting y = kx

So,

Putting the value of k

Question 30.

The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Answer:

Given slope at any point = sum of coordinates = x + y

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – 1, Q = x

I.F = e^∫Pdx

= e^{– dx}

= e ^{– x}

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × e ^{– x} = ∫ x × e ^{– x} dx + c

⇒ ye ^{– x} = ∫ x × e ^{– x} dx + c (Using integration by parts)

⇒ ye ^{– x} = – x e ^{– x –} e ^{– x} + c

⇒y = – x – 1 + ce^x……(1)

As the equation passing through origin,

0 = 0 – 1 + c× 1

⇒ c = 1

Putting the value of c in equation (1)

∴ y = – x – 1 + e^x

⇒ x + y + 1 = e^x

Question 31.

Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.

Answer:

Given slope at any point = sum of the abscissa and the product of the abscissa and the ordinate = x + xy

_{According to question,}

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – x, Q = x

I.F = e^∫Pdx

= e^{– xdx}

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × = ∫ x × dx + c……(1)

Let I = ∫ x × dx

Let

⇒

∴ I = ∫ – dt = –

Now substituting the value of I in equation (1)

⇒ y + c

⇒y = – 1 + c……(2)

As the equation passing through (0,1),

1 = – 1 + c× 1

⇒ c = 2

Putting the value of c in equation (1)

∴ y = – 1 + 2

Question 32.

The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point ( – 1, 1).

Answer:

Given the slope of the curve = square of the abscissa = x²

⇒

Integrating both sides we have,

⇒ ∫dy = ∫x²dx

⇒ ……(1)

The curve passes through point ( – 1,1)

⇒

Putting the value of c in equation (1)

Question 33.

Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the ab.

Answer:

Given product of slope of the curve and ordinate = x

⇒

Integrating both sides we have,

⇒ ∫ydy = ∫xdx

⇒ ……(1)

The curve passes through point (0,a)

⇒

Putting the value of c in equation (1)

Question 34.

The x - intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).

Answer:

Let P(x,y) be the point on the curve y = f(x) such that tangent at Pcuts the coordinate axes at A and B.

It cuts the axes at A and B so, equation of tangent at P(x,y)

Y – y = (X – x)

Now, putting Y = 0

0 – y = (X – x)

⇒ X = x – y

So, B(x – y,0)

Given, intercept on x – axis = y

⇒ x – y = y

⇒ – y = y – x

⇒

⇒ ……(1)

We can see that it is a linear differential equation.

Comparing it with

P = , Q = – 1

I.F = e^∫Pdy

= e^dy

= e ^{– logy} = y

Solution of the given equation is given by

x × I.F = ∫Q × I.F dy + c

⇒ x × = ∫ – 1 × dy + c

⇒ = – logy + c……(1)

As the equation passing through (1,1)

0 = – 1 + c

⇒ c = 1

Putting the value of c in equation (1)

∴ = – logy + 1

⇒ x = y – ylogy

Very Short Answer

Question 1.

Define a differential equation.

Answer:

Differential Equation: An equation containing independent variable, dependent variable, and differential coefficient of dependent variable with respect to independent variable.

Here y is dependent variable and x is independent variable.

Examples:

Question 2.

Define a differential equation.

Answer:

Differential Equation: An equation containing independent variable, dependent variable, and differential coefficient of dependent variable with respect to independent variable.

Here y is dependent variable and x is independent variable.

Examples:

Question 3.

Define order of a differential equation

Answer:

ORDER: The order of a differential equation is the order of the highest derivative involved in the equation.

Examples:

⇒ In example 1 order of differential equation is 1.

⇒ In example 2 order of differential equation is 2.

Question 4.

Define order of a differential equation

Answer:

ORDER: The order of a differential equation is the order of the highest derivative involved in the equation.

Examples:

⇒ In example 1 order of differential equation is 1.

⇒ In example 2 order of differential equation is 2.

Question 5.

Define degree of a differential equation

Answer:

DEGREE: The degree of differential equation is represented by the power of the highest order derivative in the given differential equation.

The differential equation must be a polynomial equation in derivatives for the degree to be defined and must be free from radicals and fractions.

Examples:

⇒ In example 1 order of differential equation is 2 and its degree is 1.

⇒ In example 2 order of differential equation is 1 and its degree is 2.

⇒ In example 3, the differential equation is not a polynomial equation in derivatives. Hence, the degree for this equation is not defined.

Question 6.

Define degree of a differential equation

Answer:

DEGREE: The degree of differential equation is represented by the power of the highest order derivative in the given differential equation.

The differential equation must be a polynomial equation in derivatives for the degree to be defined and must be free from radicals and fractions.

Examples:

⇒ In example 1 order of differential equation is 2 and its degree is 1.

⇒ In example 2 order of differential equation is 1 and its degree is 2.

⇒ In example 3, the differential equation is not a polynomial equation in derivatives. Hence, the degree for this equation is not defined.

Question 7.

Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.

Answer:

We are given

y = Cx + 5 ----(1)

Differentiating w.r.t x we get,

Put value of C in (1)

is the required differential equation.

Question 8.

Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.

Answer:

We are given

y = Cx + 5 ----(1)

Differentiating w.r.t x we get,

Put value of C in (1)

is the required differential equation.

Question 9.

Write the differential equation obtained by eliminating the arbitrary constant C in the equation x² – y² = C².

Answer:

We are given

x² – y² = C²

Differentiating w.r.t x we get,

is the required differential equation.

Question 10.

Write the differential equation obtained by eliminating the arbitrary constant C in the equation x² – y² = C².

Answer:

We are given

x² – y² = C²

Differentiating w.r.t x we get,

is the required differential equation.

Question 11.

Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C².

Answer:

We are given

xy = C²

Differentiating w.r.t x we get,

is the required differential equation.

Question 12.

Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C².

Answer:

We are given

xy = C²

Differentiating w.r.t x we get,

is the required differential equation.

Question 13.

Write the degree of the differential equation

Answer:

We are given

Here the order of differential equation is 2 and its degree is 4.

Question 14.

Write the degree of the differential equation

Answer:

We are given

Here the order of differential equation is 2 and its degree is 4.

Question 15.

Write the order of the differential equation

Answer:

We are given

Here the Order of differential equation is 2 and its Degree is 3.

Question 16.

Write the order of the differential equation

Answer:

We are given

Here the Order of differential equation is 2 and its Degree is 3.

Question 17.

Write the order and degree of the differential equation

Answer:

We are given

Squaring both sides, we get,

Here the order of differential equation is 1 and its degree is 2.

Question 18.

Write the order and degree of the differential equation

Answer:

We are given

Squaring both sides, we get,

Here the order of differential equation is 1 and its degree is 2.

Question 19.

Write the degree of the differential equation

Answer:

We are given

Here the order of differential equation is 2 and its degree is not defined as highest order derivative is a function of logarithmic function and it is not a polynomial equation in derivatives.

Question 20.

Write the degree of the differential equation

Answer:

We are given

Here the order of differential equation is 2 and its degree is not defined as highest order derivative is a function of logarithmic function and it is not a polynomial equation in derivatives.

Question 21.

Write the order of the differential equation of the family of circles touching X-axis at the origin.

Answer:

Differential Equation of the family of circles touching X-axis at the origin is

= x² + (y – k)² = k²

Here (0,k) is the center of the circle and radius k.

= x² + y² + k² – 2 × k × y = k²

= x² + y² – 2 × k × y = 0 ---(1)

Differentiate w.r.t x we get,

Put value of k in (1) we get,

is the differential equation of the family of circles touching X-axis at the origin.

Here the order of differential equation of the family of circles touching X-axis at the origin is 1.

Question 22.

Write the order of the differential equation of the family of circles touching X-axis at the origin.

Answer:

Differential Equation of the family of circles touching X-axis at the origin is

= x² + (y – k)² = k²

Here (0,k) is the center of the circle and radius k.

= x² + y² + k² – 2 × k × y = k²

= x² + y² – 2 × k × y = 0 ---(1)

Differentiate w.r.t x we get,

Put value of k in (1) we get,

is the differential equation of the family of circles touching X-axis at the origin.

Here the order of differential equation of the family of circles touching X-axis at the origin is 1.

Question 23.

Write the order of the differential equation of all non-horizontal lines in a plane.

Answer:

We know equation of a line in a plane is

ax + by = 1

Now equation of non-horizontal lines in a plane is given by,

ax + by = 1, a ≠ 0

Now a and b are two constants here we differentiate twice w.r.t y we get,

Since a ≠ 0 then,

is the differential equation of all non-horizontal lines in a plane.

Here the order of differential equation of all non-horizontal lines in a plane is 2.

Question 24.

Write the order of the differential equation of all non-horizontal lines in a plane.

Answer:

We know equation of a line in a plane is

ax + by = 1

Now equation of non-horizontal lines in a plane is given by,

ax + by = 1, a ≠ 0

Now a and b are two constants here we differentiate twice w.r.t y we get,

Since a ≠ 0 then,

is the differential equation of all non-horizontal lines in a plane.

Here the order of differential equation of all non-horizontal lines in a plane is 2.

Question 25.

If sin x is an integrating factor of the differential equation dy/dx + Py = Q, then write the value of P.

Answer:

Since is a linear differential equation

Integrating factor = e^{∫ p dx} = sin x (Given)

Taking log both sides we get,

⇒ log (e^{∫ p dx}) = log (sin x)

⇒ ∫ p dx log (e)= log (sin x)

⇒ ∫ p dx = log (sin x) ∵ log (e) = 1

Differentiate w.r.t x we get,

Question 26.

If sin x is an integrating factor of the differential equation dy/dx + Py = Q, then write the value of P.

Answer:

Since is a linear differential equation

Integrating factor = e^{∫ p dx} = sin x (Given)

Taking log both sides we get,

⇒ log (e^{∫ p dx}) = log (sin x)

⇒ ∫ p dx log (e)= log (sin x)

⇒ ∫ p dx = log (sin x) ∵ log (e) = 1

Differentiate w.r.t x we get,

Question 27.

Write the order of the differential equation of the family of circles of radius r.

Answer:

To find the Order we first need to find the differential equation of the family of circles of radius r.

In general, the equation of circle with center (a,b) and radius r is given by,

(x – a)² + (y – b)² = r² ---(1)

Differentiating above equation w.r.t x

---(2)

Again differentiating above equation w.r.t x

---(3)

Putting value of (2) and (3) in (1) we get,

is the required differential equation.

Here the order of differential equation of the family of circles of radius r is 2.

Question 28.

Write the order of the differential equation of the family of circles of radius r.

Answer:

To find the Order we first need to find the differential equation of the family of circles of radius r.

In general, the equation of circle with center (a,b) and radius r is given by,

(x – a)² + (y – b)² = r² ---(1)

Differentiating above equation w.r.t x

---(2)

Again differentiating above equation w.r.t x

---(3)

Putting value of (2) and (3) in (1) we get,

is the required differential equation.

Here the order of differential equation of the family of circles of radius r is 2.

Question 29.

Write the order of the differential equation whose solution is y = a cos x + b sin x + c e^–x.

Answer:

Solution of differential equation is

y = a cos x + b sin x + c e^–x ---(1)

Since it has 3 constants a, b, c we differentiate it by 3 times

Differentiate w.r.t x we get,

---(2)

Again, differentiate w.r.t x we get,

---(3)

Again, differentiate w.r.t x we get,

---(4)

Equation (3) implies

From (1)

---(5)

Now, adding equation (2) and (4) we get,

is the required differential equation.

Here the order of differential equation is 3.

Question 30.

Write the order of the differential equation whose solution is y = a cos x + b sin x + c e^–x.

Answer:

Solution of differential equation is

y = a cos x + b sin x + c e^–x ---(1)

Since it has 3 constants a, b, c we differentiate it by 3 times

Differentiate w.r.t x we get,

---(2)

Again, differentiate w.r.t x we get,

---(3)

Again, differentiate w.r.t x we get,

---(4)

Equation (3) implies

From (1)

---(5)

Now, adding equation (2) and (4) we get,

is the required differential equation.

Here the order of differential equation is 3.

Question 31.

Write the order of the differential equation associated with the primitive y = C₁ + C₂ e^x + C₃ e^–2x + C₄, where C₁, C₂, C₃, C₄ are arbitrary constants.

Answer:

y = C₁ + C₂ e^x + C₃ e^–2x + C₄ ---(1)

Since it has 4 arbitrary constants C₁, C₂, C₃, C₄ we differentiate it by 4 times

Differentiate w.r.t x we get,

---(2)

Again, differentiate w.r.t x we get,

---(3)

Again, differentiate w.r.t x we get,

---(4)

Again, differentiate w.r.t x we get,

---(5)

Now, (2) – (3) we get

---(6)

Now, (4) – (5) we get

From (6)

is the required differential equation

Here the order of differential equation is 4.

Question 32.

Write the order of the differential equation associated with the primitive y = C₁ + C₂ e^x + C₃ e^–2x + C₄, where C₁, C₂, C₃, C₄ are arbitrary constants.

Answer:

y = C₁ + C₂ e^x + C₃ e^–2x + C₄ ---(1)

Since it has 4 arbitrary constants C₁, C₂, C₃, C₄ we differentiate it by 4 times

Differentiate w.r.t x we get,

---(2)

Again, differentiate w.r.t x we get,

---(3)

Again, differentiate w.r.t x we get,

---(4)

Again, differentiate w.r.t x we get,

---(5)

Now, (2) – (3) we get

---(6)

Now, (4) – (5) we get

From (6)

is the required differential equation

Here the order of differential equation is 4.

Question 33.

What is the degree of the following differential equation?

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

Question 34.

What is the degree of the following differential equation?

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

Question 35.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

Question 36.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

Question 37.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

Question 38.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

Question 39.

Write the differential equation representing family of curves y = mx, where m is arbitrary constant.

Answer:

We are given the equation representing family of curves as,

y = mx --(1)

Differentiate w.r.t x we get,

Put value of m in equation (1) we get,

is the required differential equation representing family of curves y = mx.

Question 40.

Write the differential equation representing family of curves y = mx, where m is arbitrary constant.

Answer:

We are given the equation representing family of curves as,

y = mx --(1)

Differentiate w.r.t x we get,

Put value of m in equation (1) we get,

is the required differential equation representing family of curves y = mx.

Question 41.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 2

Question 42.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 2

Question 43.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 3

Question 44.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 3

Question 45.

Write degree of the differential equation

Answer:

Here Order of differential equation is 2.

Its degree is not defined as highest order derivative is a function of logarithmic function and it is not a polynomial equation in derivatives.

Question 46.

Write degree of the differential equation

Answer:

Here Order of differential equation is 2.

Its degree is not defined as highest order derivative is a function of logarithmic function and it is not a polynomial equation in derivatives.

Question 47.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2.

Its degree is not defined as it is not a polynomial equation in derivatives.

Question 48.

Write the degree of the differential equation

Answer:

Here Order of differential equation is 2.

Its degree is not defined as it is not a polynomial equation in derivatives.

Question 49.

Write the order and degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 4

Question 50.

Write the order and degree of the differential equation

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 4

Question 51.

The degree of the differential equation

Answer:

Here Order of differential equation is 2

Its degree is not defined as one derivative is exponent of exponential function and it is not a polynomial equation in derivatives

Question 52.

The degree of the differential equation

Answer:

Here Order of differential equation is 2

Its degree is not defined as one derivative is exponent of exponential function and it is not a polynomial equation in derivatives

Question 53.

How many arbitrary constants are there in the general solution of the differential equation of order 3.

Answer:

Let any differential equation of order 3 be

Here A is any constant.

Now, to know the number of arbitrary constants in the general solution we integrate both sides of equation (1)

Again integrating

is the general solution of the differential equation with 3 arbitrary constants C₁, C_2, C₃.

∴ There are 3 arbitrary constants in the general solution of the differential equation of order 3.

Question 54.

How many arbitrary constants are there in the general solution of the differential equation of order 3.

Answer:

Let any differential equation of order 3 be

Here A is any constant.

Now, to know the number of arbitrary constants in the general solution we integrate both sides of equation (1)

Again integrating

is the general solution of the differential equation with 3 arbitrary constants C₁, C_2, C₃.

∴ There are 3 arbitrary constants in the general solution of the differential equation of order 3.

Question 55.

Write the order of the differential equation representing the family of curves y = ax + a³.

Answer:

We are given

y = ax + a³ -- (1)

Differentiating w.r.t x we get

Put value of a in equation (1) we get,

is the required differential equation.

Since order is the highest order derivative present in the differential equation.

∴ Order = 1 and Degree = 3.

Question 56.

Write the order of the differential equation representing the family of curves y = ax + a³.

Answer:

We are given

y = ax + a³ -- (1)

Differentiating w.r.t x we get

Put value of a in equation (1) we get,

is the required differential equation.

Since order is the highest order derivative present in the differential equation.

∴ Order = 1 and Degree = 3.

Question 57.

Find the sum of the order and degree of the differential equation

Answer:

Order = Highest order derivative present in the differential equation.

∴ Order = 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

∴ Sum of the order and degree = 2 + 1 = 3

Question 58.

Find the sum of the order and degree of the differential equation

Answer:

Order = Highest order derivative present in the differential equation.

∴ Order = 2

Degree = Highest power of highest order derivative which is

∴ Degree = 1

∴ Sum of the order and degree = 2 + 1 = 3

Question 59.

Find the solution of the differential equation

Answer:

We can write above differential equation as

By the method of variable separable we can write,

Integrating both sides,

Let 1 + y² = t and 1 + x² = u

⇒ 2y dy = dt ⇒ 2x dx = du

Putting values in integral we get,

Putting values of t and u,

Where C is the arbitrary constant.

is the required solution of the differential equation.

Question 60.

Find the solution of the differential equation

Answer:

We can write above differential equation as

By the method of variable separable we can write,

Integrating both sides,

Let 1 + y² = t and 1 + x² = u

⇒ 2y dy = dt ⇒ 2x dx = du

Putting values in integral we get,

Putting values of t and u,

Where C is the arbitrary constant.

is the required solution of the differential equation.

Mcq

Question 1.

Mark the correct alternative in each of the following:

The integrating factor of the differential equation is given by

A. log (log x)

B. e^x

C. log x

D. x

Answer:

Dividing x.log x both sides we get,

The above equation is of the form i.e. linear differential equation.

Integrating factor = e^{∫ P}^dx

Considering ∫ P

Put log x = t

Putting values, we get,

∵ e^{log x} = x

∴ I.F = log x = (C)

Question 2.

Mark the correct alternative in each of the following:

The general solution of the differential equation is:

A. log y = kx

B. y = kx

C. xy = k

D. y = k log x

Answer:

Given:

By the method of Variable separable we get,

Taking integral both sides

⇒log y=log x + log k

where log k is an arbitrary constant.

⇒log y=log x . k

∵ log x + log y = log xy

⇒y = kx

=(B)

is the required general solution.

Question 3.

Mark the correct alternative in each of the following:

Integrating factor of the differential equation is

A. sin x

B. sec x

C. tan x

D. cos x

Answer:

Dividing cos x both sides we get,

The above equation is of the form i.e. linear differential equation.

Integrating factor = e^{∫ P}^dx

Considering ∫ P

=sec x

∵ e^{log x} = x

∴ I.F = sec x = (B)

Question 4.

Mark the correct alternative in each of the following:

The degree of the differential equation is

A. 1/2

B. 2

C. 3

D. 4

Answer:

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 2

= (B)

Question 5.

Mark the correct alternative in each of the following:

The degree of the differential equation is

A. 4

B. 2

C. 5

D. 10

Answer:

Cubing both sides, we get,

Order = Highest order derivative present in the differential equation.

Here Order of differential equation is 2

Degree = Highest power of highest order derivative which is

∴ Degree = 3

Question 6.

Mark the correct alternative in each of the following:

The general solution of the differential equation is

A. x + y sinx = C

B. x + y cosx = C

C. y + x (sin x + cos x) = C

D. y sin x = x + C

Answer:

The above equation is of the form i.e. linear differential equation.

Here, P = cot x and Q = cosec x

Integrating factor = e^{∫ P}^dx

Considering ∫ P

∵ e^{log x} = x

∴ I.F = sin x

Now, General solution is

⇒ y (I.F) = ∫ Q (I.F) dx + C

⇒ y sin x = ∫ cosec x sin x dx + C

⇒y sin x = x + C

=(D)

Question 7.

Mark the correct alternative in each of the following:

The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is

A. yⁿ + y’ = 0

B. yⁿ – ω²y = 0

C. yⁿ = –ω²y

D. yⁿ + y = 0

Answer:

y = A cos ωt + B sin ωt ---(1)

Since we need to eliminate A and B, so we differentiate (1) twice.

Differentiating (1) w.r.t t we get,

Again differentiating (1) w.r.t t we get,

⇒y'' = -ω² y

=(C)

Question 8.

Mark the correct alternative in each of the following:

The equation of the curve whose slope is given by x > 0, y > 0 and which passes through the point (1, 1) is

A. x² = y

B. y² = x

C. x² = 2y

D. y² = 2x

Answer:

Slope of the curve is given by

By Variable separable

Integrating both sides

∵ log x + log y = log xy

-- (1)

Since curve passes through the point (1, 1), we get

⇒√1=C.1

⇒C=1

Putting value of C in (1) we get,

⇒x=√y

Squaring both sides

⇒x² = y

=(A)

Question 9.

Mark the correct alternative in each of the following:

The order of the differential equation whose general solution is given by

y = c₁cos (2x + c₂) – (c₃ + c₄) a^x^{+ c5} + c₆ sin (x – c₇) is

A. 3

B. 4

C. 5

D. 2

Answer:

y = c₁cos (2x + c₂) – (c₃ + c₄) a^x^{+ c5} + c₆ sin (x – c₇)

⇒ y = c₁ [cos(2x). cos c₂ – sin (2x). sin c �₂] – (c₃ + c₄) a^c5. a^x + c₆[sin (x). cos c₇ – cos (x). sin c₇)

⇒ y = c₁.cos c_{2 .} cos(2x)– c₁. sin c �₂. sin (2x)– (c₃ + c₄) a^c5. a^x +c₆. cos c_{7 � �}. sin (x) – c₆. sin c_7.cos (x)

Now, c₁.cos c₂,c₁. sin c �₂, (c₃ + c₄) a^c5, c₆. cos c_{7 � �}, c₆. sin c₇ are all constants

∴ c₁.cos c₂ = A

c₁. sin c �₂ = B

(c₃ + c₄) a^c5 = C

c₆. cos c₇ = D

c₆. sin c₇ = E

⇒ y = A. cos(2x)– B. sin (2x)– C. a^x +D _{� �}. sin (x) – E. cos (x)

Where A, B, C, D and E are constants

Since there are 5 constants, we have to differentiate y w.r.t x five times.

So, the Order of the differential equation = 5

= (C)

Question 10.

Mark the correct alternative in each of the following:

The solution of the differential equation represents a circle when

A. a = b

B. a = –b

C. a = –2b

D. a = 2b

Answer:

By Variable separable,

⇒ (by + f) dy = (ax + g) dx

Integrating both sides

⇒ ∫ (by + f) dy = ∫ (ax + g) dx

⇒ by² + 2fy = ax² + 2gx + C^’

Where C’ = 2C

We know general solution of circle is

(x – h)² + (y – k)² = r²

Where (h, k) is center of the circle and r is radius.

By completing square method

Now to form above equation as equation of circle we need

i.e. a = -b

= (B)

Question 11.

Mark the correct alternative in each of the following:

The solution of the differential equation with y(1) = 1 is given by

A.

B.

C.

D.

Answer:

By the method of variable separable

Integrating both sides we get,

⇒log √y=log C. x^-1

∵ log a + log b = log ab

Squaring both sides, we get,

Now the given condition is y(1) = 1

⇒C=1

∴ The solution of the differential equation is

=(A)

Question 12.

Mark the correct alternative in each of the following:

The solution of the differential equation is given by

A. y = xe^x+C

B. x = ye^x

C. y = x + C

D. xy e^x + C

Answer:

By the method of variable separable

Integrating both sides we get,

∴ The solution of the differential equation is

Question 13.

Mark the correct alternative in each of the following:

The order of the differential equation satisfying is

A. 1

B. 2

C. 3

D. 4

Answer:

The given curve is

Since the number of constants in the given curve is 1 i.e. a which is an arbitrary constant.

Also, Number of arbitrary constants in the equation of the curve = Order of the differential equation of the curve.

∴ Order = 1

= (A)

Question 14.

Mark the correct alternative in each of the following:

The solution of the differential equation y₁y₃ = y₂² is

A. x = C₁ e^{C2 y} + C₃

B. y = C₁ e^{C2 x} + C₃

C. 2x = C₁ e^{C2 y} + C₃

D. none of these

Answer:

y₁y₃ = y₂²

We can write above differential equation as

⇒ y' y’’’ = (y’’)²

Integrating both sides we get,

Again, integrating both sides we get,

∴ The solution of the differential equation is

=(B)

Question 15.

Mark the correct alternative in each of the following:

The general solution of the differential equation where g(x) is a given function of x, is

A. g(x) + log {1 + y + g(x)} = C

B. g(x) + log {1 + y – g(x)} = C

C. g(x) – log {1 + y – g(x)} = C

D. none of these

Answer:

Since it is a form of linear differential equation where

P = g’(x) and Q = g(x) g’(x)

Integrating Factor (I.F) = e^{∫ p dx}

I.F = e^{∫ g’(x) dx} = e^g(x)

Solution of differential equation is given by

y.(I.F) = ∫ Q.(I.F) dx + C

⇒ y. e^g(x) = ∫ g(x). g’(x). e^g(x) dx + C

Consider integral ∫ g(x). g’(x). e^g(x) dx

Put g(x) = t

⇒ g’(x) dx = dt

⇒ ∫ t. e^t dt

Treating t as first function and e^t as second function, So integrating by Parts we get,

⇒ t. e^t - ∫ 1.e^t dt + C

⇒ e^t (t – 1) + C

Putting value of t we get,

⇒ e ^g(x) (g(x) – 1) + C

∴ y. e ^g(x) = e ^g(x) (g(x) – 1) + C

Dividing e ^g(x) both sides we get,

⇒ y = (g(x) – 1) + C e^-g(x)

⇒ y - g(x) + 1 = C e^-g(x)

Taking log both sides we get,

⇒ log (y – g(x) + 1) = log (C e^-g(x))

⇒ log (y – g(x) + 1) = log C – g(x) log e

⇒ log (y – g(x) + 1) = log C – g(x) ∵ log e = 1

⇒ g(x) + log {1 + y – g(x)} = C ⇒ (B) where log C = C

Question 16.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A.

B.

C. y = tan (C + x + x²)

D.

Answer:

By the method of variable separable

Integrating both sides we get,

Now the given condition is y(0) = 0

Question 17.

Mark the correct alternative in each of the following:

The differential equation of the ellipse is

A.

B.

C.

D. none of these

Answer:

Since the equation has 2 constants so we differentiate twice,

Differentiating w.r.t x

--(1)

Again, differentiating w.r.t x

--(2)

Substitute value of (2) in (1) we get,

Dividing both sides by y.y’ we get,

Question 18.

Mark the correct alternative in each of the following:

Solution of the differential equation is

A. x (y + cos x)= sin x + C

B. x (y – cos x)= sin x + C

C. x (y + cos x)= cos x + C

D. none of these

Answer:

Since it is a form of linear differential equation.

Integrating Factor (I.F) = e^{∫ p dx}

Solution of differential equation is given by

y.(I.F) = ∫ Q.(I.F) dx + C

⇒ y. x = ∫ (sin x).x dx + C

Consider integral ∫ (sin x).x dx

Treating x as first function and sin x as second function. So, integrating by Parts we get,

⇒ x. (-cos x) + ∫ 1.cos x dx + C

⇒ – x. cos x + sin x + C

∴ y. x = – x. cos x + sin x + C

⇒ x (y + cos x) = sin x + C = (A) is the required solution.

Question 19.

Mark the correct alternative in each of the following:

The equation of the curve satisfying the differential equation

y(x + y³)dx = x(y³ – x) dy and passing through the point (1, 1) is

A. y³ – 2x + 3x²y = 0

B. y³ + 2x + 3x²y = 0

C. y³ + 2x – 3x²y = 0

D. none of these

Answer:

y(x + y³)dx = x(y³ – x)dy

⇒ yx dx + y⁴ dx = xy³ dy – x² dy

⇒ xy³ dy – x² dy – yx dx – y⁴ dx = 0

⇒ y³ [x dy – y dx] – x[x dy + y dx] = 0

Divide both sides by y²x³ we get,

Integrating both sides we get,

-- (1)

Now the given curve is passing through the point (1, 1)

Substituting value of C in (1) we get,

⇒ y³ + 2x = 3x²y

∴ y³ + 2x – 3x²y = 0 = (C) is the required solution.

Question 20.

Mark the correct alternative in each of the following:

The solution of the differential equation represents

A. circles

B. straight lines

C. ellipses

D. parabolas

Answer:

By the method of variable separable we get,

Integrating both sides,

∵ log a + log b = log ab

Squaring both sides, we get,

Where A = C²

Since it is the form of (y – k)² = 4p(x – h) which represents parabolas.

i.e. (y – (-3))² = A(x – 0)

∴ The solution of the differential equation represents parabolas = (D)

Question 21.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A.

B.

C.

D.

Answer:

--(1)

The above equation is of the form of Homogeneous differential equation.

Differentiate w.r.t x we get,

--(2)

Putting value of (2) in (1) we get

Integrating both sides,

Putting value of v we get,

Question 22.

Mark the correct alternative in each of the following:

The differential equation satisfied by ax² + by² = 1 is

A. xyy₂ + y₁² + yy₁ = 0

B. xyy₂ + xy₁² – yy₁ = 0

C. xyy₂ – xy₁² + yy₁ = 0

D. none of these

Answer:

ax² + by² = 1

Since it has two arbitrary constants, we differentiate it twice w.r.t x

Differentiate w.r.t x

--(1)

Again, differentiate w.r.t x

Applying product rule

Put value of a in (1) we get,

Which is ⇒ xyy₂ + xy₁² – yy₁ = 0 =(B)

Question 23.

Mark the correct alternative in each of the following:

The differential equation which represents the family of curves y = e^Cx is

A. y₁ = C₂ y

B. xy₁ – ln y = 0

C. x ln y = yy₁

D. y ln y = xy₁

Answer:

y = e^Cx

Taking log both sides we get,

⇒ log y = log e^Cx

⇒ log y = Cx log e ∵ log a^x = x log a

⇒ log y = Cx --(1) ∵ log e = 1

Differentiate w.r.t x we get,

Put value of C in (1) we get,

Which is ⇒ y ln y = xy₁ = (D)

Question 24.

Mark the correct alternative in each of the following:

Which of the following transformations reduce the differential equation into the form

A. u = log x

B. u = e^x

C. u = (logz)^–1

D. u = (logz)²

Answer:

Dividing z(log z)² both sides we get,

--(1)

Put (log z)^-1 = u

Differentiate w.r.t x

Putting values in (1) we get,

The above equation is of the form

So the required substitution is u = (logz)^–1 = (C)

Question 25.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A.

B.

C.

D.

Answer:

--(1)

The above differential is homogeneous differential equation with degree 0

Differentiate w.r.t x we get,

--(2)

Putting value of (2) in (1) we get

Integrate both sides we get,

Question 26.

Mark the correct alternative in each of the following:

If m and n are the order and degree of the differential equation then

A. m = 3, n = 3

B. m = 3, n = 2

C. n = 3, n = 5

D. m = 3, n = 1

Answer:

Order = Highest order derivative present in the differential equation.

∴ Order = 3 = m

Degree = Highest power of highest order derivative which is

∴ Degree = 2 = n

∴ m = 3, n = 2 = (B)

Question 27.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A. (x + y) e^x+y = 0

B. (x + C) e^x+y = 0

C. (x – C) e^x+y = 1

D. (x – C) e^x+y + 1 = 0

Answer:

Put x + y = z

Differentiating w.r.t x we get,

Substituting values, we get

Integrating both sides

Putting value of z, we get

Question 28.

Mark the correct alternative in each of the following:

The solution of is

A. x² + y² = 12x + C

B. x² + y² = 3x + C

C. x³ + y³ = 3x + C

D. x³ + y³ = 12x + C

Answer:

By variable separable we get,

Integrating both sides

⇒ y³ = 12x – x³ + C Where 3k = C

⇒ x³ + y³ = 12x + C = (D)

Question 29.

Mark the correct alternative in each of the following:

The family of curves in which the subtangent at any point of a curve is double the abscissae, is given by

A. x = Cy²

B. y = Cx²

C. x² = Cy²

D. y = Cx

Answer:

We are given,

The family of curves in which the subtangent at any point of a curve is double the abscissae

By variable separable we get,

Integrating both sides

∵ log a + log b = log ab

Squaring both sides, we get

⇒ x = Cy² = (A) Where K² = C

Question 30.

Mark the correct alternative in each of the following:

The solution of the differential equation x dx + y dy = x²y dy – y² x dx, is

A. x² – 1 = C (1 + y²)

B. x² + 1 = C (1 – y²)

C. x³ – 1 = C (1 + y³)

D. x² + 1 = C (1 – y³)

Answer:

x dx + y dy = x²y dy – y² x dx

⇒ x dx + y² x dx = x²y dy – y dy

⇒ x dx(1 + y²) = y dy(x² – 1)

By Variable separable

Integrating both sides we get

-- (1)

Put x² – 1 = t and Put 1 + y² = u

Diff w.r.t x Diff w.r.t y

2x dx = dt 2y dy = du

Putting values in (1) we get,

∵ log a + log b = log ab

Putting values of t and u we get,

⇒ x² – 1 = C (1 + y²) = (A)

Question 31.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A. y = 2 + x²

B.

C. y = x(x –1)

D.

Answer:

By variable separable we get,

Integrating both sides we get,

Where tan C = C

Now since C is arbitrary constant, we put C = 1 (let) we get,

Question 32.

Mark the correct alternative in each of the following:

The differential equation has the general solution

A. y – x³ = 2cx

B. 2y – x³ = cx

C. 2y + x² = 2cx

D. y + x² = 2cx

Answer:

Divide both sides by x we get,

Since it is a form of linear differential equation where

and Q = x²

Integrating Factor (I.F) = e^{∫ p dx}

Solution of differential equation is given by

y.(I.F) = ∫ Q.(I.F) dx + C

⇒ 2y – x³ = cx = (B)

Question 33.

Mark the correct alternative in each of the following:

The solution of the differential equation approaches to zero when x → ∞, if

A. k = 0

B. k > 0

C. k < 0

D. none of these

Answer:

By variable separable and integrating both sides

Where e^c = A

We have given condition y(0) = 1

We know e^∞ → ∞ and e^-^∞ → 0

So, we are given that y approaches to zero when x → ∞

i.e. 0 = e^k^∞

Which is possible only when k < 0 = (C)

Question 34.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A. tan^–1 x – tan^–1 y = tan^–1 C

B. tan^–1 y – tan^–1 x = tan^–1 C

C. tan^–1 y ± tan^–1 x = tanC

D. tan^–1 y + tan^–1 x = tan^–1 C

Answer:

By variable separable we get,

Integrating both sides we get,

⇒ tan^–1 y + tan^–1 x = tan^–1 C = (D)

Question 35.

Mark the correct alternative in each of the following:

The solution of the differential equation is

A.

B.

C.

D.

Answer:

--(1)

The above differential is homogeneous differential equation

Differentiate w.r.t x we get,

--(2)

Putting value of (2) in (1) we get

By variable separable and integrate both sides we get,

Putting value of v, we get

Question 36.

Mark the correct alternative in each of the following:

The differential equation n > 2 can be reduced to linear form by substituting

A. z = y^n-1

B. z = yⁿ

C. z = yⁿ⁺¹

D. z = y^1-n

Answer:

Multiply both sides by y^-n

-- (1)

Put y^1-n = z

Differentiate w.r.t x we get,

--(2)

Put value of (2) in (1)

Multiply both sides by (1 – n)

Since it is a form of linear differential equation

∴ we put z = y^1-n to form linear differential equation = (D)

Question 37.

Mark the correct alternative in each of the following:

If p and q are the order and degree of the differential equation then

A. p < q

B. p = q

C. p > q

D. none of theses

Answer:

Order = Highest order derivative present in the differential equation.

∴ Order = 2 = p

Degree = Highest power of highest order derivative which is

∴ Degree = 1 = q

∴ p = 2 > q = 1 = (C)

Question 38.

Mark the correct alternative in each of the following:

Which of the following is the integrating factor of ?

A. x

B. e^x

C. log x

D. log (log x)

Answer:

Divide both sides by x log x we get,

Since it is a form of linear differential equation where

and

Integrating Factor (I.F) = e^{∫ p dx}

Put log x = t

Differentiate w.r.t x we get,

Putting value in integral

Putting value of t, we get,

Now,

Question 39.

Mark the correct alternative in each of the following:

What is integrating factor of ?

A. sec x + tan x

B. log (sec x + tan x)

C. e^{sec x}

D. sec x

Answer:

Since it is a form of linear differential equation where

and

Integrating Factor (I.F) = e^{∫ p dx}

Question 40.

Mark the correct alternative in each of the following:

Integrating factor of the differential equation is

A. cos x

B. tan x

C. sec x

D. sin x

Answer:

Divide both sides by cos x we get,

Since it is a form of linear differential equation where

and

Integrating Factor (I.F) = e^{∫ p dx}

Question 41.

Mark the correct alternative in each of the following:

The degree of the differential equation is

A. 3

B. 2

C. 1

D. not defined

Answer:

Here Order of differential equation is 2.

Its degree is not defined as it is not a polynomial equation in derivatives. = (D)

Question 42.

Mark the correct alternative in each of the following:

The order of the differential equation is

A. 2

B. 1

C. 0

D. not defined

Answer:

Order = Highest order derivative present in the differential equation which is

∴ Order = 2 = (A)

Question 43.

Mark the correct alternative in each of the following:

The number of arbitrary constants in the general solution of differential equation of fourth order is

A. 0 B. 0

C. 3 D. 4

Answer:

We know,

the number of arbitrary constants of an ordinary differential equation (ODE) is given by the order of the highest derivative.

∵ differential equation is of fourth order then it will have 4 arbitrary constants in the general solution. = (D)

Question 44.

Mark the correct alternative in each of the following:

The number of arbitrary constants in the particular solution of a differential equation of third order is

A. 3

B. 2

C. 1

D. 0

Answer:

We know,

the number of arbitrary constants of an ordinary differential equation (ODE) is given by the order of the highest derivative and if we give particular values to those arbitrary constants, we get particular solution in which we have 0 arbitrary constants

∴ The number of arbitrary constants in the particular solution of a differential equation of third order is 0 = (D)

Question 45.

Mark the correct alternative in each of the following:

Which of the following differential equations has y = C₁ e^x + C₂ e^–x as the general solution?

A.

B.

C.

D.

Answer:

Solving for (A)

For general solution put (D² + 1) = 0

General solution, y = C₁ cos x + C₂ sin x

Solving for (B)

For general solution put (D² – 1²) = 0

General solution, y = C₁ e^x + C₂ e^-x = (B) which is required solution.

Question 46.

Mark the correct alternative in each of the following:

Which of the following differential equations has y = x as one of its particular solution?

A.

B.

C.

D.

Answer:

y = x

Differentiate w.r.t x we get,

Consider,

∵ x = y

Question 47.

Mark the correct alternative in each of the following:

The general solution of the differential equation is

A. e^x + e^–y = C

B. e^x + e^y = C

C. e^–x + e^y = C

D. e^–x + e^–y = C

Answer:

By variable separable and integrating both sides we get,

⇒ e^x + e^–y = C = (A)

Question 48.

Mark the correct alternative in each of the following:

A homogeneous differential equation of the form can be solved by making the substitution.

A. y = vx

B. v = yx

C. x = vy

D. x = v

Answer:

--(1)

For solving the above homogeneous differential equation we must put

--(2)

We put value of (2) in (1) for finding the solution.

Question 49.

Mark the correct alternative in each of the following:

Which of the following is a homogeneous differential equation?

A. (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0

B. xy dx – (x³ + y³) dy = 0

C. (x³ + 2y²) dx + 2xy dy = 0

D. y² dx + (x² – xy) – y²) dy = 0

Answer:

We know the property of homogeneous differential equation i.e.

⇒ f (λ x, λ y) = f (x, y)

In the given set of options, option (D) is correct as addition of power is same throughout the equation.

Question 50.

Mark the correct alternative in each of the following:

The integrating factor of the differential equation

A. e^–x

B. e^–y

C. 1/x

D. x

Answer:

Divide both sides by x we get,

Since it is a form of linear differential equation where

and

Integrating Factor (I.F) = e^{∫ p dx}

Question 51.

Mark the correct alternative in each of the following:

The general solution of the differential equation is

A. xy = C

B. x = Cy²

C. y = Cx

D. y = Cx²

Answer:

By variable separable and integrating both sides we get,

⇒ y = Cx = (C)

Question 52.

Mark the correct alternative in each of the following:

The general solution of a differential equation of the type is

A.

B.

C.

D.

Answer:

Since it is a form of linear differential equation.

Where, P = P₁ and Q = Q₁

Integrating Factor (I.F) = e^{∫ p dy}

I.F = e^{∫ P}₁^dy

Solution of differential equation is given by

x.(I.F) = ∫ Q.(I.F) dy + C

Question 53.

Mark the correct alternative in each of the following:

The general solution of the differential equation e^x dy + (y e^x + 2x)dx = 0 is

A. x e^y + x² = C

B. x e^y + y² = C

C. y e^x + x² = C

D. y e^y + x² = C

Answer:

e^x dy + (y e^x + 2x)dx = 0

We can write above equation as

Divide both sides by e^x we get,

Since it is a form of linear differential equation.

Integrating Factor (I.F) = e^{∫ p dx}

I.F = e^{∫ 1 dx} = e^x

Solution of differential equation is given by

y.(I.F) = ∫ Q.(I.F) dx + C

⇒ y e^x + x² = C = (C)

Exercise 22.2

Question 1.

Form the differential equation of the family of curves represented by y² = (x – c)³.

Answer:

y² = (x – c)³

On differentiating the above equation with respect to x we get

Putting the value of (x – c) in the given equation, we get,

On squaring, both sides we get,

Hence, is the differential equation which represents the family of curves y² = (x – c)³.

Question 2.

Form the differential equation corresponding to y = e^mx by eliminating m.

Answer:

Given equation, y = e^mx

On differentiating the above equation with respect to x we get

But y = e^mx

Now we have, y = e^mx

Applying log on both sides, we get,

log y = mx

which gives

So, putting this value of m in we get

Hence, is the differential equation corresponding to y = e^mx.

Question 3.

Form the differential equation from the following primitives where constants are arbitrary:

y² = 4ax

Answer:

On differentiating with respect to x, we get

On substituting the value of a we get,

Hence, is the differential equation corresponding to

y² = 4ax.

Question 4.

Form the differential equation from the following primitives where constants are arbitrary:

y = cx + 2c² + c³

Answer:

On differentiating with respect to x, we get,

Putting this value of c in the given equation we get

Hence, is the differential equation corresponding to y = cx + 2c² + c³.

Question 5.

Form the differential equation from the following primitives where constants are arbitrary:

xy = a²

Answer:

Again, differentiating with respect to x we get,

Hence, is the differential equation corresponding to xy = a².

Question 6.

Form the differential equation from the following primitives where constants are arbitrary:

y = ax² + bx + c

Answer:

As the given equation has 3 different arbitrary constants so we can differentiate it thrice with respect to x

So, differentiating once with respect to x,

Differentiating twice with respect to x,

Now, differentiating thrice with respect to x we get,

Hence, is the differential equation corresponding to

y = ax² + bx + c.

Question 7.

Form the differential equation of the family of curves y = Ae^2x + Be^–2x, where A and B are arbitrary constants.

Answer:

y = Ae^2x + Be^–2x

As the equating has two different arbitrary constants so, we can differentiate it twice with respect to x. So, on differentiating once with respect to x we get,

Again, differentiating it with respect to x, we get

But, Ae^2x + Be^–2x = y (Given)

Hence the differential equation corresponding to the curves

y = Ae^2x + Be^–2x is

Question 8.

Form the differential equation of the family of curves,

x = A cos nt + B sin nt, where A and B are arbitrary constant.

Answer:

As the given equation has two different arbitrary constants so we can differentiate it twice with respect to x.

x = A cos nt + B sin nt

On differentiating with respect to t we get,

Again, differentiating with respect to x,

As x = A cos nt + B sin nt

Hence, is the required differential equation.

Question 9.

Form the differential equation corresponding to y² = a(b – x²) by eliminating a and b.

Answer:

Given equation y² = a(b – x²)

On differentiating with respect to x, we get,

Again, differentiating with respect to x we get,

From (1) we have

On putting, this value in (2) we get,

So, the required differential equation is .

Question 10.

Form the differential equation corresponding to y² – 2 ay + x² = a² by eliminating a.

Answer:

y² – 2 a y + x² = a²

On differentiating, with respect to x we get,

Putting this value of a in the given equation, we get,

⇒ y²y'² – 2y²y'² – 2xyy' + x²y'² = y² y'² + 2xyy' + x²

⇒ y²y'² – 2y²y'² – 2xyy' + x²y'² – y² y'² – 2xyy' – x² = 0

⇒ – 4xyy' + y'²x² – x² – 2y'²y² = 0

⇒ y’²(x² – 2y²) – 4xyy’ – x² = 0

So, y’²(x² – 2y²) – 4xyy’ – x² = 0

Question 11.

Form the differential equation corresponding to (x – a)² + (y – b)² = r² by eliminating a and b.

Answer:

(x – a)² + (y – b)² = r² …… (i)

On differentiating with respect to x, we get,

Again, differentiating with respect to x we get,

Put the value of (y – b) obtained in (ii) we get,

Put the value of (x – a) and (y – b) in (i) we get,

Put we get,

⇒ (y’³ + y’)² + (y’² + 1)² = r²y’’²

So, the required differential equation is (y’³ + y’)² + (y’² + 1)² = r²y’’².

Question 12.

Form the differential equation of all the circles which pass through the origin and whose centers lie on the y – axis.

Answer:

Any circle with centre at (h, k) and radius r is given by,

(x – h)² + (y – k)² = r²

Here centre is on y – axis, so h = 0

So, we have the equation of circle as, x² + (y – k)² = r²

Further, it is given that circle passes through the origin (0,0) therefore origin must satisfy the equation of circle. So, we get,

0 + k² = r²

So, the equation of circle is x² + (y – k)² = k²

⇒ x² + y² – 2ky = 0

⇒ x² + y² = 2ky

Now, differentiating it with respect to x we get,

Hence, the required differential equation is

Question 13.

Find the differential equation of all the circles which pass through the origin and whose centers lie on the x - axis.

Answer:

Any circle with centre at (h, k) and radius r is given by,

(x – h)² + (y – k)² = r²

Here centre is on x - axis, so k = 0

So, we have the equation of circle as, (x – h)² + y² = r²

Further it is given that circle passes through origin (0,0) therefore origin must satisfy equation of circle. So, we get,

0 + h² = r²

So, the equation of circle is (x – h)² + y² = h²

⇒ x² – 2hx + y² = 0

⇒ x² + y² = 2hx

Now, differentiating it with respect to x we get,

Hence, the required differential equation is

Question 14.

Assume that a raindrop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the raindrop.

Answer:

Let r be the radius of the raindrop, V be its volume and A be its surface area

Given

Negative because V decreases with an increase in t

is a proportionality constant

Now, we know that

So, we have,

Hence, the required differential equation is

Question 15.

Find the differential equation of all the parabolas with latus rectum ‘4a' and whose axes are parallel to the x - axis.

Answer:

Equation of parabola with latus rectum ‘4a’ and axes parallel to x - axes and vertex at (h,k) is given by

(y – k)² = 4a(x – h)

On differentiating with respect to x we get,

Again differentiating (i) with respect to x we get,

From (i) we have , on substituting it in the above equation we get,

Hence, the required differential equation is

Question 16.

Show that the differential equation of which is a solution, is

Answer:

On differentiating with respect to x we have,

Now,

= 4x³

Which is the given equation.

Hence, is solution to the given differential equation.

Question 17.

Form the differential equation having y = (sin^–1 x)² + A cos^–1 x + B, where A and B are arbitrary constants, as its general solution.

Answer:

On differentiating with respect to x we get,

Again, differentiating with respect to x we have,

Hence the required differential equation is

Question 18.

Form the differential equation of the family of curves represented by the equation (a being the parameter):

i. (2x + a)² + y² = a²

ii. (2x – a)² – y² = a²

iii. (x – a)² + 2y² = a²

Answer:

(i)

(2 x + a)² + y² = a²

On differentiating, with respect to x we have,

Putting this value of a in the given equation we get,

ii. (2 x – a)² – y² = a²

⇒ 4x² + a² – 4ax – y² = a²

⇒ 4x² – 4ax – y² = 0

⇒ 4ax = 4x² – y²

On differentiating with respect to x we get,

iii. (x – a)² + 2 y² = a²

On differentiating, with respect to x we have,

Putting this value of a in the given equation we get,

Question 19.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

x² + y² = a²

Answer:

On differentiating we get,

Question 20.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

x² – y² = a²

Answer:

On differentiating we get,

Question 21.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

y² = 4ax

Answer:

On differentiating we get,

Question 22.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

x² + (y – b)² = 1

Answer:

On differentiating we get,

Put (ii) in (i),

Question 23.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

(x – a)² – y² = 1

Answer:

Differentiating with respect to x we get,

Now putting the value of (x – a) in the initial equation, we get

Question 24.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

Answer:

b²x² – a²y² = a²b²

On differentiating with respect to x we get,

Again, differentiating with respect to x we get,

Putting this value of b² in (i) we get,

Question 25.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

y² = 4 a (x – b)

Answer:

On differentiating with respect to x

Again, differentiating with respect to x we get,

Question 26.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

y = ax³

Answer:

y = ax³

On differentiating with respect to x we get,

From the given equation

So, we have

Question 27.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

x² + y² = ax³

Answer:

x² + y² = ax³

Differentiating with respect to x,

Question 28.

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):

y = e^ax

Answer:

Differentiating with respect to x

From the given equation we have,

y = e^ax

= >log y = ax

Now,

Question 29.

Form the differential equation representing the family of ellipses having the center at the origin and foci on the x - axis.

Answer:

Equation of required ellipse is

Where a,b are arbitrary constants

Differentiating (i) with respect to x we get,

Now, differentiating (ii) with respect to x we get,

The required differential equation is

Question 30.

Form the differential equation of the family of hyperbolas having foci on x - axis and center at the origin.

Answer:

Equation of required hyperbola is

Where a,b are arbitrary constants

Differentiating with respect to x we get,

Again, differentiating with respect to x we get,

Substituting this value of in (ii) we get,

The required differential equation is

Question 31.

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Answer:

Let C denote the family of circles in the second quadrant and touching the coordinate axes and let ( – a, a) be co - ordinate of the centre of any member of this circle

Now, the equation representing this family of circle is (x + a)² + (y – a)² = a² …… (i)

⇒ x² + y² + 2ax – 2ay + a² = 0 …… (ii)

Differentiating (ii) with respect to x we get,

Substituting this value of a in (i) we get,

The required differential equation is

Exercise 22.3

Question 1.

Show that y = be^x + ce^2x is a solution of the differential equation,

Answer:

The differential equation is and the function that is to be proven as solution is

y = be^x+ ce^2x, now we need to find the values of and .

be^x + 2ce^2x

be^x + 4ce^2x

Putting the values of these variables in the differential equation, we get,

be^x + 4ce^2x – 3(be^x + 2ce^2x) + 2(be^x + ce^2x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied. Hence, this function is the solution of the differential equation.

Question 2.

Verify that y = 4 sin 3x is a solution of the differential equation

Answer:

The differential equation is and the function that is to be proven as the solution is

y = 4 sin 3x, now we need to find .

12 cos 3x

– 36 sin 3x

Putting the values in the equation, we get,

–36 sin 3x + 9(4 sin 3x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 3.

Show that y = ae^2x + be^–x is a solution of the differential equation

Answer:

The differential equation is and the function that is to be proven as solution is

y = ae^2x + be^–x, now we need to find the value of and .

= 2ae^2x – be^–x

= 4ae^2x + be^–x

Putting these values in the equation, we get,

4ae^2x + be^–x –(2ae^2x – be^–x) – 2(ae^2x + be^–x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 4.

Show that the function y = A cos x + B sin x is a solution of the differential equation

Answer:

The differential equation is and the function that is to be proven as solution is

y = A cos x + B sin x, now we need to find the value of .

= –A sin x + B cos x

= –A cos x – B sin x

Putting the values in equation, we get,

–A cos x – B sin x + A cos x + B sin x = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

Question 5.

Show that the function y = A cos2x – B sin 2x is a solution of the differential equation

Answer:

The differential equation is and the function that is to be proven as solution is

y = A cos2x – B sin 2x, now we find the value of .

= –2A sin 2x – 2B cos 2x

= –4A cos 2x + 4B sin 2x

Putting the values in the equation, we get,

–4A cos 2x + 4B sin 2x + 4(A cos 2x – B sin 2x) = 0,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 6.

Show that y = Ae^Bx is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is y = Ae^Bx, now we need to find the value of and .

= ABe^Bx

= AB²e^Bx

Putting values in the equation,

L.H.S. = R.H.S.

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 7.

Verify that is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is , now we need to find the value of and .

Putting values in equation,

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 8.

Verify that y² = 4ax is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is y² = 4ax, now we need to find the value of and .

Putting the values,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 9.

Show that Ax² + By² =1 is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is Ax² + By² =1, now we need to find the value of and .

Putting the values in the equation,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 10.

Show that y = ax³ + bx² + c is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is

y = ax³ + bx² + c; now we need to find the value of .

Putting the value of variables in the equation,

6a = 6a

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

Question 11.

Show that is a solution of the differential equation= 0.

Answer:

The differential equation is and the function to be proven is the solution of equation is , now we need to find the value of .

Putting the value of the variables in the equation,

1 + c² x² + 2 c x – 1 – c² – x² – c² x² + c² + x² – 2 c x = 0

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 12.

Show that y = e^x (A cosx + B sinx) is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is

y = e^x (A cosx + B sinx), we need to find the value of .

= e^x(A cos x + B sin x) + e^x(–A sin x + B cos x)

= e^x(A cos x + B sin x) + e^x(–A sin x + B cos x) + e^x(–A sin x + B cos x) + e^x(–A cos x – B sin x)

= 2e^x(–A sin x + B cos x)

Putting the values in equation,

2e^x(–A sin x + B cos x) – 2e^x(A cos x + B sin x) – 2e^x(–A sin x + B cos x) + 2 e^x(A cos x + B sin x) = 0

0 = 0

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

Question 13.

Verify that y = cx + 2c² is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is

y = cx + 2c², now we need to find the value of .

= c + 0

Putting the values,

2c² + xc – cx – 2c² = 0

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 14.

Verify that y = –x – 1 is a solution of the differential equation (y – x)dy – (y² – x²)dx = 0.

Answer:

The differential equation is and the function to be proven as the solution is

y = – x – 1, now we need to find the value of .

= –1

Putting the values in equation,

–1 = –x –1 +x

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 15.

Verify that y² = 4a(x + a) is a solution of the differential equation

Answer:

The differential equation is and the function to be verified as the solution is

y² = 4a(x+a), now we need to find the value of .

Putting the value in equation,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 16.

Verify that is a solution of the differential equation

Answer:

The differential equation is and the function to be verified as the solution is , now we need to find the value of and .

Putting the values,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 17.

Verify that is a solution of the differential equation.

Answer:

The differential equation is and the function to be verified as the solution is , now we need to find the value of and .

Putting the values in the equation,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 18.

Verify that is a solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is , now we need to find the value of and .

Putting the values in the equation,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 19.

Show that the differential equation of which is a solution is

Answer:

The differential equation is and the function to be proven as the solution is , now we need to find the value of .

Putting the value,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 20.

Show that y = e^–x + ax + b is solution of the differential equation

Answer:

The differential equation is and the function to be proven as the solution is

y = e^–x + ax + b, now we need to find the value of .

= –e^–x + a

= e^–x

Putting the values in equation,

(e^x) (e^–x) = 1

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Question 21.

For each of the following differential equations verify that the accompanying function is a solution.

Differential equation

i.

ii.

iii.

iv.

v.

Function

Answer:

(i). The differential equation isand the function to be proven as solution is y = ax, now we need to find the value of .

= a

Putting the value,

ax = y = ax,

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

(ii). The differential equation is and the function to be proven as the solution of this equation is , now we need to find the value of .

Putting the values,

x – x = 0

As, L.H.S = R.H.S. the equation is satisfied, hence this function is the solution of the differential equation.

(iii). The differential equation is and the function to be proven as solution is

now we need to find the value of .

As L.H.S. ≠ R.H.S. the equation is not satisfied, hence this function is not the solution of this differential equation.

(iv). The differential equation is and the function to be proven as solution is , now we need to find the value of .

Putting the values,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

(v). The differential equation is and the function to be proven as solution is , now we need to find the value of .

Putting the value, we get,

As, L.H.S = R.H.S. the equation is satisfied, so hence this function is the solution of the differential equation.

Exercise 22.4

Question 1.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y = log x

Answer:

Verification:

y = log x

Differentiating both sides we get,

Multiplying x on both the sides

Also, at x=1, y should be equal to 0. Let’s check it out.

At x=1, y=log (1)=0. (Hence the initial value condition is also satisfied)

Question 2.

For each of the following initial value problems verify that the accompanying function is a solution:

Function:

Answer:

Verification:

Differentiating both sides we get,

Since y=e^x , we can replace e^x in the above differential equation

Hence y = e^x is the solution of the differential equation.

Also, at x=0, we get y=e⁰ which is equal to 1.

Question 3.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y=sin x

Answer:

Verification:

y=sin x

Differentiating both sides we get,

Therefore, sin x is the solution to the differential equation.

Also at x=0, we get y=sin 0 which is equal to 0.

Question 4.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y=e^x+1

Answer:

Verification:

Y = e^x + 1

Differentiating both sides we get,

Therefore, e^x+1 is the solution of the differential equation.

Also at x=0, we get y=e⁰+1 which is equal to 2.

Question 5.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y = e^-x+2

Answer:

Verification:

y = e^-x + 2

Differentiating both sides we get,

Therefore, e^-x + 1 is the solution of the differential equation.

Also at x = 0, we get y = e⁰ + 2 which is equal to 3.

Question 6.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y = sin x + cos x

Answer:

y = sin x + cos x

Differentiating both sides we get,

Therefore, sin x + cos x is the solution of the differential equation.

Also at x=0, we get y=sin 0+ cos 0 which is equal to 0+1=1.

Question 7.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y = e^x + e^-x

Answer:

y = e^x + e^-x

Differentiating both sides we get,

Therefore, e^x + e^-x is the solution of the differential equation.

Also at x=0, we get y=e⁰ + e⁰ which is equal to 1+1=2.

Question 8.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y = e^x+e^2x

Answer:

Verification:

y = e^x + e^2x

Differentiating both sides we get,

Therefore, e^x + e^2x is the solution of the differential equation.

Also at x=0, we get y=e⁰ + e⁰ which is equal to 1+1=2.

Also at x=0, we get y¹=e⁰+2e⁰=3.

Question 9.

For each of the following initial value problems verify that the accompanying function is a solution:

Function: y=xe^x + e^x

Answer:

Verification:

Y = x e^x + e^x

Differentiating both sides we get,

Therefore, e^x + xe^x is the solution of the differential equation.

Exercise 22.5

Question 1.

Solve the following differential equations

Answer:

By Separate the variables

Integrate both side

as we known

Question 2.

Solve the following differential equations

Answer:

By Separate the variables

Integrate both side

as we known

Question 3.

Solve the following differential equations

Answer:

Integrate both sides we get,

as we known &

Question 4.

Solve the following differential equations

Answer:

Integrate both sides we get,

we known that

Question 5.

Solve the following differential equations

Answer:

We known that

by identity

Integrate both sides we get,

we know that

Question 6.

Solve the following differential equations

Answer:

On dividing

Integrate both sides we get,

we know

Question 7.

Solve the following differential equations

Answer:

Integrate both side

Now using integration by parts we get,

Let

Differentiate with respect to x.

Put value of t=1+x²

Question 8.

Solve the following differential equations

Answer:

Integrate both sides we get,

Now integrating by parts we get,

Question 9.

Solve the following differential equations

Answer:

Integrate both sides we get,

Since,

[Using Integration by parts]

]

Question 10.

Solve the following differential equations

Answer:

Integrate both sides we get,

We know that

Let

Differentiate with respect to x.

put the value of t

we know that

Question 11.

Solve the following differential equations

(sinx + cosx)dy + (cosx – sinx) dx = 0

Answer:

By separating variables

Integrate both sides we get,

Let

Differentiate with respect to x.

Put t value in above eq.

Question 12.

Solve the following differential equations

Answer:

Separate variables

Integrate both sides we get,

…1

logx=t

Differentiate with respect to x.

Put value of t

…2

we know that:

Using Integration by parts we get,

…3

Put values of eq 2 and 3 in eq 1^st

Question 13.

Solve the following differential equations

Answer:

By separating variables

Integrate both sides we get,

Let

Differentiate with respect to x.

Using integration by parts we get,

Put z = x³

Question 14.

Solve the following differential equations

Answer:

We know the trigonometric identity and

Separate variables and Integrate both sides,

Let

Differentiate with respect to x.

put z = cosec x

Question 15.

Solve the following differential equations

Answer:

By identity:

Separate variables

Integrate both sides we get,

Question 16.

Solve the following differential equations

Answer:

Integrate both sides we get,

Let

we know that:

Put the value of t,

Question 17.

Solve the following differential equations

Answer:

Integrate both sides we get,

Question 18.

Solve the following differential equations

Answer:

Separate variables

Integrate both sides we get,

Let

Differentiate with respect to x,

Put value of t and z

Question 19.

Solve the following differential equations

Answer:

Integrate both sides we get,

Integrating by parts we get,

Question 20.

Solve the following differential equations

Answer:

Separate variables

Integrate both sides we get,

Question 21.

Solve the following differential equations

Answer:

Integrate both sides we get,

…1

Solv in partial fraction we get,

By solving we get

a = 1/2

B = 1

C = -1/2

Let x² = t

X dx = dt/2

Put the value of t

Question 22.

Solve the following initial value problem:

Answer:

Integrate both sides we get,

At x=0 , y=1

Therefore,

So, c = 1

Putting c = 1 in above eq. we get,

Question 23.

Solve the following initial value problem:

Answer:

Taking log

Integrate both sides we get,

Using Integration by parts we get,

At x = 0, y = 3

Put on above eq.

Question 24.

Solve the following initial value problem:

Answer:

Separation

Integrate both side

At x=0 c=100

Put in above eq.

Question 25.

Solve the following initial value problem:

Answer:

Separate variabless

Integrate both sides we get,

At x = -1, y = 0

Put in above eq.

Question 26.

Solve the following initial value problem:

Answer:

Separate variables,

Integrate both sides we get,

Let

Put value of t,

At x=2 y=0

Put in above eq.

Exercise 22.6

Question 1.

Solve the following differential equations:

Answer:

Separate variables

Integrate both sides we get,

Let y²=t

Differentiate with respect to x

Put value of t,

Question 2.

Solve the following differential equations:

Answer:

Separate variables

Integrate both sides we get,

Let y²=t

Differentiate with respect to x

Put the value of t,

Question 3.

Solve the following differential equations:

Answer:

Separate variables

We know that

Integrate both sides we get,

Question 4.

Solve the following differential equations:

Answer:

By separate variables,

Integrate both sides we get,

Using trigonometry identity:

Exercise 22.7

Question 1.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on other side, we have

Integrating LHS with respect to y and RHS with respect to x

Adding 2 and subtracting 2, to the numerator of RHS

Re - writing RHS as

Using identities:

and

Integrating both sides, we have

log(y) = 2x + 2log(x – 1) + c

Question 2.

Solve the following differential equations:

(1 + x²)dy = (xy)dx

Answer:

Now separating variable x on one side and variable y on other side, we have

Integrating both sides

Using identities:

and for RHS assuming x^{2 =} t (substitution property) and differentiating both sides

Now, 2xdx = dt

Substituting the above value in the integral and replacing x² with t and integrating both sides

Now replacing t by x²

Taking anti - log both sides

y² = 1+x²

Question 3.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Using identities:

and

Integrating both sides we get,

log(y) = e^x + x + c

Question 4.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Adding 1 and subtracting 1 to the numerator of RHS

Using formula: x³ – 1 = (x – 1) (x² + 1 + x)

Using identities:

and

And integrating both sides we get,

Question 5.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Now integrating both sides we get,

Using identities:

and

Question 6.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identities:

and

Question 7.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identities:

and for RHS using property

Sin(y) = e^x(log(x)) + c

Question 8.

Solve the following differential equations:

Answer:

Re - writing the question as:

Integrating both sides using identities:

log(y – 1) – log(y) = log(x) + c

using:

Taking anti - log both sides we have,

Question 9.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on other side, we have

Using identities:

and on RHS side assuming e^y = t, so e^ydy = dt by differentiating both sides.

Now integrating both sides

– log |sin(x)| = log(t+1) + c

Replacing t by e^y

– log |sin(x)| = log(e^y+1) + c

[sin(x)] (e^y+1) = c

Question 10.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using integration by parts method.

According to integration by parts method,

Using identity:

Question 11.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Re – writing LHS as

Integrating both sides using identities:

and

y – log(y – 1) = x + log(x) + c

y – x = x(y – 1) + c

Question 12.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identities:

and

log(|sec(y)|) = – log(x) + c

using log(a)+log(b) = log(ab) formula, we have,

x sec(y) = c

Question 13.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identities:

and property

Question 14.

Solve the following differential equations:

Answer:

Re - writing the question as

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identities:

and

Question 15.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Re - writing the equation as

Now assuming 1+y² = t²

Differentiating both sides, we get

ydy = tdt

Similarly, for LHS assuming 1+x² = v²

differentiating both sides

xdx = vdv

substituting these values in the differential equation

Integrating both sides

Re - writing as

Using identity:

and

Integrating both sides, we get

Substituting the value of v and t in the above equation

Question 16.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Adding 1 and subtracting 1 to the numerator of LHS, we get

Integrating both sides using identities:

And

Assuming e^x = t and differentiating both sides we get,

e^xdx = dt

substituting this value in above equation

y – log(y+1) = log(1+t)

substituting t as e^x

y – log(y+1) = log(1+e^x) + c

Question 17.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identity:

Question 18.

Solve the following differential equations:

Answer:

Re - writing the equation as

Now separating variable x on one side and variable y on another side, we have

Multiplying and dividing the numerator of LHS by x

Assuming 1+y² = t² and 1+x² = v²

Differentiating we get,

ydy = tdt

xdx = vdv

substituting these values in above differential equation

Integrating both sides

Adding 1 and subtracting 1 to the numerator of RHS

Using identities:

and

Substituting t and v in above equation

Question 19.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have,

y(2log(y) + 1) dy = e^x(sin²x + sin2x) dx

Integrating both sides we get,

Using integration by parts for LHS and identity:

for RHS.

Question 20.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have,

(sin y + y cos y) dy = (2 x log x + x) dx

Integrating both sides we get,

Using identity:

and integration by parts we get,

ysiny = x²log(x) + c

Question 21.

Solve the following differential equations:

Answer:

Re - writing the equation as (1 – x²) dy = xdx(y² – y)

Now separating variable x on one side and variable y on another side, we have

Integrating both sides

Using identity:

and assuming x² = t

Differentiating both sides we get,

2 x dx = dt

Substituting this value in above equation

Replacing t by x²

Question 22.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have,

– cot(x) dx = sec²y cot(y) dy

Integrating both sides we get,

Sin2x = 2 sinx cosx

Using identities:

and

– log| sinx | = log| tany |+ c

Using: log(a)+log(b) = log(ab)

log( sin x tan y) = log c

sin(x) tan(y) = c

Question 23.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

Integrating both sides

Using identity:

and substituting x² = t and y² = v

x² = t

2xdx = dt

Similarly, for y

y² = v

2ydy = dv

Substituting these values in above integral equation

Substituting the values of t and v in above equation

Question 24.

Solve the following differential equations:

Answer:

Using the formula:

Re - writing the equation as

Now separating variable x on one side and variable y on another side, we have

Sec y ta ny dy = 2 sin x dx

Integrating both sides using identities :

And

sec(y) = – 2cos(x) + c

Question 25.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

coty dy = – tanx dx

Integrating the above equation using identities:

and

log|sin(y)| = – log|cos(y)|+c

Using: log(a)+log(b) = log(ab)

sin(y) cos(y) = c

Question 26.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have

– cot(y)dy = cos(x)dx

Integrating both sides using identities :

and

– log|sin(y)| = sin(x) + c

Question 27.

Solve the following differential equations:

Answer:

Now separating variable x on one side and variable y on another side, we have,

Let 1 – x² = t² and 1 – y² = t²

Differentiating we get

xdx = – tdt

ydy = – vdv

substituting these values in above differential equation

dt = – dv

integrating both sides, we get

t = – v + c

substituting the value of v and c in above equation

Question 28.

Solve the following differential equations:

Answer:

Re – writing the equation as

y(1+x) dx+x(1 – y²) dy = 0

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identity:

And

Using log(a)+log(b) = log(ab)

Question 29.

Solve the following differential equations:

Answer:

Re - writing the above equation as

Now separating variable x on one side and variable y on another side, we have

Integrating both sides using identities:

And

Question 30.

Solve the following differential equation:

Answer:

Given. = 0

Find: Find the general solution of this differential equation.

= C

Hence, The solution of the given Differential Equation is C

Question 31.

Solve the following differential equation:

dy + (x + 1)(y + 1) dx = 0

Answer:

_{Given. dy + (x + 1)(y + 1)dx = 0}

Find: Find the general solution of this differential equation.

_{= dy = -(x + 1)(y + 1)dx}

= C

Hence, The solution of the given differential equation is C.

Question 32.

Solve the following differential equation:

Answer:

Given

Find: Find the general solution of this differential equation.

= C

= + C

Hence, The solution of the given differential equation is + C

Question 33.

Solve the following differential equation:

Answer:

Given

Find: Find the general solution of this differential equation.

Integrate Both side

Question 34.

Solve the following differential equation:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

Integrate Both Side,

= C

Hence, The solution of the given differential equation is C

Question 35.

Solve the following differential equation:

Answer:

Given differential Equation

Find: Find the general solution of this differential equation.

Integrate Both side,

= C

Hence,the solution is C

Question 36.

Solve the following differential equation:

Answer:

Given

Find: Find the general solution of this differential equation.

Multiply by 2 Both side

Now, Integrate both sides,

= Let assume y^{2 +} 2 = t Let assume x² + 2 = v

Then 2y dy =dt 2x dx = dv

= log|t| = -log|v|

Put the value of t and v

Hence,

Question 37.

Solve the following differential equation:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

Integrating Both side

Using integration by parts both side

Hence, The Solution is C

Question 38.

Solve the following differential equation:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

= y – log|y + 1| = log|x| + x + log|C|

= y = log|x| + x + log|y + 1| + log|C|

Hence, y = log|cx(y + 1)| + x

Question 39.

Solve the following differential equation:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

= C = (1 – x²)(1 + y²)

Question 40.

Solve the following differential equation:

Answer:

Given

Find: Find the general solution of this differential equation.

= Integrating on the both side we get,

Hence,

Question 41.

Solve the following differential equation:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

= C

Question 42.

Solve the following initial value problem:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

Integrating both sides,

= .....(i)

Put x =0, y = 2

2 ⟹

2 = C

Put value of C in (i)

Hence, y =

Question 43.

Solve the following initial value problem:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

= .....(i)

Put x =1, y = 2

= C = 4

Put C =4 in equation (i)

Hence,

Question 44.

Solve the following initial value problem:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

On Integrating we get,

= y – 2log|y + 2|=log |x| + log |C| ….(i)

Put y = 0, x = 2

= 0 – 2log2 = log 2 + log c

= -2 log 2 –log 2 = log C

= -3 log 2 = log c

= =log c

Put the value of C in equation (i)

Hence, y – 2log|y + 2|=

Question 45.

Solve the following initial value problem:

Answer:

Given

Integrating both side

= C .....(i)

Put x = 0, y = 1/2

= -2 = 2 + c

C = -4

Put the value of C = -4 in equation (i)

= - 4

Hence,

Question 46.

Solve the following initial value problem:

Answer:

Given differential equation

Find: Find the general solution of this differential equation.

Integrating both side

= C .....(i)

Put t =0, r = r₀ in equation (i).

Now, C

= C

Put the value of C in eq(i)

Hence,

Question 47.

Solve the following initial value problem:

Answer:

Given differential equation

Find: Find the particular solution of this differential equation.

Integrating both side

= + C

Put y =1 and x=1

= + C

= 0 = C

= C =

So, + C

Hence, y =

Question 48.

Solve the following initial value problem:

Answer:

The given differential equation is

Find: Find the particular solution of this differential equation.

Integrating both sides,

= log|y|=log|sec x| + log|C| .....(I)

Put y =1, x=0

0 = log(1) + C

C = 0

Put the value of C in equation in equation(I)

= log y = log |sec x|

Hence, y = sec x

Question 49.

Solve the following initial value problem:

Answer:

The given differential Equation is .

Find: Find the particular solution of this differential equation.

Now Integrating Both sides

= …..(1)

Put x = 1, y = 1

= 2 log(1)=5log(1) + C

= 0 = C

Put the value of C in equation (1)

= 2log |y| = 5log|x|

= y⁵ = |x|²

Hence, y = |x|^5/2

Question 50.

Solve the following initial value problem:

Answer:

The Given equation is

Find: Find the particular solution of this differential equation.

Integrating both side,

= C .....(i)

Put y = -1, x = 0

1 = e^{0 +} C

1 = 1 + C

C = 0

Put the value of C in equation (i)

Hence, y =

Question 51.

Solve the following initial value problem:

Answer:

Find: Find the particular solution of this differential equation.

On equating we get,

= cos y dy = e^x dx

Integrating both side

= sin y = e^x + C …..(i)

Put x = 0, y = π/2

= 1 = 1 + C

= C = 0

Put the value of C in equation (i)

Sin y = e^x

Hence, y = sin^-1(e^x)

Question 52.

Solve the following initial value problem:

Answer:

Find: Find the particular solution of this differential equation.

Integrating both sides

= C .....(i)

Put x = 0, y = 1

log(1) = 0 + C

0 = 0 + C

C = 0

Put the value of C in equation (i)

Hence,

Question 53.

Solve the following initial value problem:

Answer:

Given :

Find: Find the particular solution of this differential equation.

Integrating both side

= ….(i)

Put x = 0, y =1

Put the value of C in equation (i)

Hence,

Question 54.

Solve the following initial value problem:

Answer:

Given:

Find: Find the particular solution of this differential equation.

= y – 2 log|y + 2| = x + 2 log|x| + C

= y – 2 log|y + 2| - x - 2 log|x| = C

Put x =1, y = -1

= -1 -1-2log(-1 + 2) – 2log 1 = C

= -2 = C

Thus, we have

Hence, Y – x – 2log(y + 2) - 2log x = -2

Question 55.

Solve the following initial value problem:

when

Answer:

Given:

= C .....(i)

Put y = 0 and x = 0, then

= C

= C = 0

Putting the value of C in equation (i) we get.

Hence,

Question 56.

Solve the following initial value problem:

Answer:

Given:

Find: Find the particular solution of this differential equation.

Integrating both sides,

= .....(i)

Put x = 1 and y= -2 in equation (i), we get

= 2 log(1) = y + 3 – 3log(-2 + 3) + C

= 0 = 1- 0 + C

= C = -1

Put the value of C in equation (i), we get

Hence,

Question 57.

Solve the differential equation given that when

Answer:

Given differential equation.

Find: Find the particular solution of this differential equation.

Integrating both sides, we get

= log|sec y| = - log|x| + C …..(i)

Put x = , y =

= log|sec | = - log|| + C

= C

= C = log2

Put C in equation (i)

= log|sec y| = - log|x| + log2

Hence,

Question 58.

Solve the differential equation given that when

Answer:

Given:

Find: Find the particular solution of this differential equation.

Integrating both side,

= ….(i)

Put x = 0, y = 1

= C

Put C in eq (i), we get

= y + xy =1- x

Hence, x + y = 1 – xy

Question 59.

Solve the differential equation given that when

Answer:

Given:

Find: Find the particular solution of this differential equation.

Integrating both side

= C

Put y = 0, x = 1

= 0 = 0 + 1/2 + C

= C =

Put C = in equation (i)

Hence, The Solution is .

Question 60.

Find the particular solution of given that when

Answer:

Given

Find: Find the particular solution of this differential equation.

Integrating both sides

Using Integration by parts

= C ….(i)

Put y =3 and x = 0

= 3 = 0-0 + C

= C = 3

Put C = 3 in equation (i)

Hence, The Solution is 3.

Question 61.

Find the solution of the differential equation given that when

Answer:

The given differential Equation is .

Find: Find the solution of this differential equation.

= .

Integrating both side,

= log |sin y| = - sin x + C .....(i)

Put y = and x =

= log |sin | = - sin + C

= 0 = -1 + C

= C =1

Put the value of C in eq(i)

Hence, The solution is log |sin y| + sin x =1

Question 62.

Find the particular solution of the differential equation given that when

Answer:

The given differential equation is

Find: Find the solution of this differential equation.

Integrating both sides

= C

Put y =1 and x =0

= -1 = 0 + C

= C = -1

Put the value of C in eq (i)

Question 63.

Find the equation of a curve passing through the point (0, 0) and whose differential equation is

Answer:

The given differential equation

Find: Find the solution of this differential equation.

Integrating both sides

= I =

Question 64.

For the differential equation Find the solution curve passing through the point (1, -1).

Answer:

The given differential equation is

Find: Find the solution of this differential equation.

Integrating both side

= C

= .....(i)

Put x =1 and y = -1

= - 2 - C = 0

= C = -2

Put the value of C in equation (i)

Hence, The solution of the curve is .

Question 65.

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Answer:

Let the rate of change of the volume of the balloon be k (where k is constant)

Find: Find the radius of the balloon after t second.

Integrating both sides, we get:

= .....(i)

Now, at t = 0, r = 3.

= C = 36

At t = 3, r = 6:

= k = 84π

Substitute the value of K and C in equation (1), we get

= r³ = 63t + 27

= r = (63t + 27)^1/3

Hence, the radius of the balloon after t seconds is (63 t + 27)^1/3 .

Question 66.

In a bank principal increases at the rate of r% per year. Find the value of r if ` 100 double itself in 10 years (log_e 2 = 0.6931).

Answer:

Let p, t, and r represent the principal, time and rate of interest respectively.

It is the given that the principal increases continuously at the rate r% per year.

Find: Find the value of r?

Integrating both sides, we get:

= p = .....-(i)

It is given that when t = 0, p = 100

= 100 = e^k .....--(2)

Now, if t =10,then p = 2x100 =200

= from (2)

= r = 6.931

Hence, the value of r is 6.93 %.

Question 67.

In a bank principal increases at the rate of 5% per year. An amount of ` 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).

Answer:

Let p, and t represent the principal, time respectively.

It is the given that the principal increases continuously at the rate of 5% per year.

Integrating both sides, we get:

= p = .....-(i)

It is given that when t = 0, p = 1000

= 1000 = e^c .....--(2)

Now,

Putting t = 10,we get

= p = 1000 x 1.648

= p = 1648

Question 68.

In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.

Answer:

Let y be the number of bacteria at any instant t

It is given that the rate of growth of the bacteria is proportional to the number present.

= (where k is a constant)

Integrating both sides, we get

= log y = kt + C

Let y₀ be the number of bacteria at t = 0.

= log y₀ = C

Substitute the value of C in, we get

⟹ log y = kt + log y₀

= log y - log y₀ = kt

Also, it is given that the number of bacteria increased by 10% in 2 hours.

Substituting the value,

= k =

Therefore,

Now, the time when the number of bacteria increases from 100000 to 200000 be t1.

= at t = t₁

Now, =

Hence, in hours the number if bacteria increases from 100000 to 200000.

Question 69.

If y(x) is a solution of the differential equation and y(0) = 1, then find the value of y(π/2).

Answer:

Consider the given equation

Integrating both sides,

= log(1 + y) ⟹-log(2 + sin x) + log C

= log(1 + y) + log(2 + sin x) = log C

= log(1 + y)(2 + sin x) = log C

= (1 + y)(2 + sin x) = c ….(1)

Given that y(0) = 1

= (1 + 1)(2 + sin 0) = c

= C = 4

Substituting the value of C in eq (1), we get

= (1 + y)(2 + sin x) = 4

= (1 + y) =

= y = ……(2)

Now, find the value of y(π/2)

Substituting the value of x = in equation (2)

= y =

Question 70.

Find the particular solution of the differential equation given that y = 0 when
x = 1.

Answer:

Consider the differential equation

Let 1-y² = t then -2y dy = dt Let log x = v

= .....(1)

Substitute the value of v and t in eq (2)

= …..(2)

Put x =1 and y =3 in eq (2)

= C =

Put the value of C in eq (2)

Hence, The particular solution is

Exercise 22.8

Question 1.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒ ……(1)

Let us assume z = x + y + 1

Differentiating w.r.t x on both the sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same (i.e, variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that and

Also ∫adx = ax + C

⇒

⇒ tan^–1z = x + C

We know that z = x + y + 1 , substituting this we get,

⇒ tan^–1(x + y + 1) = x + C

∴ The solution for the given Differential equation is tan^–1(x + y + 1) = x + C

Question 2.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒

⇒ ……(1)

Let us assume z = x – y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same side (i.e., variable seperable technique) we get,

⇒

We know that cos2z = cos²z – sin²z = 2cos²z – 1 = 1 – 2sin²z.

⇒

We know 1 + cot²x = cosec²x

⇒

Integrating on both sides we get,

⇒

We know that:

(1) ∫cosec²x = –cotx + C

(2)

(3) ∫adx = ax + C

⇒

Since z = x – y substituting this we get,

⇒

∴ The solution for the given Differential equation is .

Question 3.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒ ……(1)

Let us assume z = x – y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same side(i.e., variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that:

(1) ∫adx = ax + C

(2)

⇒ 2z – log(z + 3) = x + C

Since z = x – y, we substitute this,

⇒ 2(x – y) – log(x–y + 3) = x + C

⇒ 2x – 2y –log(x–y + 3) = x + C

⇒ x – 2y –log(x–y + 3) = C

∴ The solution for the given Differential equation is: x – 2y –log(x–y + 3) = C.

Question 4.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒ ……(1)

Let us assume z = x + y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing the like variables to same side (i.e., Variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that:

(1)

(2)

⇒

⇒ tan^–1z = x + C

Since z = x + y we substitute this,

⇒ tan^–1(x + y) = x + C

⇒ x + y = tan(x + C)

∴ The solution for the given Differential equation is x + y = tan(x + C).

Question 5.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒

⇒ ……(1)

Let us assume z = x + y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same side (i.e., Variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that:

(1) ∫adx = ax + C

(2)

⇒

⇒ z – tan^–1z = x + C

Since z = x + y, we substitute this,

⇒ x + y – tan^–1(x + y) = x + C

⇒ y – tan^–1(x + y) = C

∴ The solution for the given Differential equation is y – tan^–1(x + y) = C.

Question 6.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒

We know that 1–cos²x = sin²x

⇒ ……(1)

Let us assume z = x– 2y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substitute (2) in (1) we get,

⇒

Bringing like variables on same side (i.e., variable seperable technique) we get,

⇒

We know that

⇒ sec²zdz = dx

Integrating on both sides we get,

⇒ ∫sec²zdz = ∫dx

We know that:

(1) ∫sec²xdx = tanx + C

(2) ∫adx = ax + C

⇒ tanz = x + C

Since z = x – 2y we substitute this,

⇒ tan(x–2y) = x + C

∴ The solution for the given Differential Equation is tan(x–2y) = x + C.

Question 7.

Solve the following differential equations:

Answer:

Given Differential Equation is:

⇒ ……(1)

Let us assume z = x + y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same side(i.e, variable seperable technique) we get,

⇒

We know that

⇒

We know that cos2z = cos²z – sin²z = 2cos²z – 1

⇒

We know that 1 + tan²x = sec²x

⇒

Integrating on both sides we get,

⇒

We know that:

(1) ∫sec²xdx = tanx + C

(2) ∫adx = ax + C

⇒

Since z = x + y, we substitute this,

⇒

∴ the solution for the given differential equation is .

Question 8.

Solve the following differential equations:

Answer:

Given Differential Equation is:

⇒ ……(1)

Let us assume z = x + y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting(2) in (1) we get,

⇒

Bringing like variables on same side(i.e., variable seperable technique) we get,

⇒

We know that

⇒

Integrating on both sides we get,

⇒

We know that:

(1)

(2) ∫adx = ax + C

⇒ z + log(cosz + sinz) = 2x + C

Since z = x + y, we substitute this,

⇒ x + y + log(cos(x + y) + sin(x + y)) = 2x + C

⇒ y + log(cos(x + y) + sin(x + y)) = x + C

∴ The solution for the given Differential Equation is y + log(cos(x + y) + sin(x + y)) = x + C.

Question 9.

Solve the following differential equations:

(x + y)(dx–dy) = dx + dy

Answer:

Given Differential equation is:

⇒ (x + y)(dx–dy) = dx + dy

⇒ (x + y)dx –(x + y)dy = dx + dy

⇒ (x + y–1)dx = (x + y + 1)dy

⇒ ……(1)

Let us assume z = x + y

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same side(i.e., variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that:

(1) ∫ adx = ax + C

(2)

⇒

Since z = x + y we substitute this,

⇒ x + y + log(x + y) = 2x + C

⇒ y + log(x + y) = x + C

∴ The solution for the given Differential equation is y + log(x + y) = x + C.

Question 10.

Solve the following differential equations:

Answer:

Given Differential Equation is :

⇒

⇒ ……(1)

Let us assume z = x + y + 1

Differentiating w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substituting (2) in (1) we get,

⇒

Bringing like variables on same side (i.e., variable seperable technique) we get,

⇒

Integrating on both sides we get,

⇒

We know that:

(1) ∫adx = ax + C

(2)

⇒ z – log(z + 1) = x + C

Since z = x + y we substitute this,

⇒ x + y–log(x + y + 1) = x + C

⇒ y–log(x + y + 1) = C

⇒ y = log(x + y + 1) + C

∴ The solution for the given Differential Equation is y = log(x + y + 1) + C.

Question 11.

Solve the following differential equations:

Answer:

Given Differential equation is:

⇒ ……(1)

Let us assume z = x + y

Differentiate w.r.t x on both sides we get,

⇒

⇒ ……(2)

Substitute(2) in (1) we get,

⇒

Bringing like variables on same side (i.e., variable seperable technique) we get,

⇒

⇒ e^–zdz = dx

Integrating on both sides we get,

⇒ ∫e^–zdz = ∫dx

We know that:

(1) ∫adx = ax + C

(2)

⇒

⇒ –e^–z = x + C

⇒ x + e^–z + C = 0

Since z = x + y we substitute this,

⇒ x + e^{–(x + y)} + C = 0

∴ The solution for the given Differential Equation is x + e^{–(x + y)} + C = 0.

Exercise 22.9

Question 1.

Solve the following equations:

x²dy + y(x + y)dx = 0

Answer:

Let us write the given differential equation in the standard form:

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation, and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that

⇒

Bringing the like variables on one side

⇒

We know that:

∫and

Integrating on both sides we get

⇒

(∵ logC is also an arbitrary constant)

⇒

(∵)

(∵ xloga = loga^x)

Applying exponential on both sides, we get,

⇒

Squaring on both sides we get,

⇒

Since y = vx

we get

⇒

Cross multiplying on both sides we get,

⇒ yx² = c²(y + 2x)

∴ The solution to the given differential equation is yx² = c²(y + 2x)

Question 2.

Solve the following equations:

Answer:

Given Differential equation is :

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume:

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation, and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that:

⇒

Bringing like variables on one side we get,

⇒

We know that:

and Also,

Integrating on both sides, we get,

⇒

(∵ LogC is an arbitrary constant)

⇒

(∵)

Since y = vx,

we get

⇒

(∵ xloga = loga^x)

⇒

(Assuming log(c²) = K a constant)

∴ The solution to the given differential equation is log(y² + x²) + 2tan^-1= K

Question 3.

Solve the following equations:

Answer:

Given differential equation can be written as:

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume:

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that:

⇒

Bringing like variables on one side we get,

⇒

We know that:

Integrating on both sides, we get,

⇒

⇒ log(v² + 1) = -logx + logC (∵ LogC is an arbitrary constant)

Since y = vx,

we get

⇒

(∵ )

Applying exponential on both sides, we get,

⇒

Cross multiplying on both sides we get,

⇒ y² + x² = Cx

∴ The solution for the given differential equation is y² + x² = Cx.

Question 4.

Solve the following equations:

Answer:

Give Differential equation is:

⇒

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume:

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that

⇒

Bringing like coefficients on same sides we get,

⇒

We know that ∫adx = ax + C and

Also,

Integrating on both sides, we get,

⇒

⇒ v = logx + C

Since y = vx,

we get,

⇒

Cross multiplying on both sides we get,

⇒ y = xlogx + Cx

∴ The solution for the given differential equation is y = xlogx + Cx

Question 5.

Solve the following equations:

(x² – y²)dx – 2xydy = 0

Answer:

Given differential equation is:

⇒ (x² – y²)dx – 2xydy = 0

⇒ (x² – y²)dx = 2xydy

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that:

⇒

Bringing Like variables on same sides we get,

⇒

We know that:

Integrating on both sides, we get,

⇒

(∵ logC is an arbitrary constant)

Multiplying with -3 on both sides we get,

⇒ log|1-3v²| = -3logx + 3logC

⇒

(∵ )

⇒

(∵ alogx = logx^a)

⇒

Applying exponential on both sides we get,

⇒

Since y = vx, we get,

⇒

Cross multiplying on both sides we get,

⇒ x(x² – 3y²) = c³

⇒ x³ – 3xy² = K (say any arbitrary constant)

∴ The solution for the differential equation is x³ – 3xy² = K

Question 6.

Solve the following equations:

Answer:

Given differential equation is:

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume:

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that:

⇒

Bringing like variables on same side we get,

⇒

We know that:

and

Also,

Integrating on both sides, we get,

⇒

Since y = vx, we get,

⇒

(∵)

⇒

(∵ alogx = logx^a)

⇒

∴ The solution for the given Differential equation is

Question 7.

Solve the following equations:

Answer:

Given Differential equation is:

⇒

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume:

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation, and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that

⇒

Bringing like variables on same side we get,

⇒

We know that:

⇒

⇒ -log(1-v²) = logx + logC

⇒ log(1-v²)^-1 = log(Cx)

(∵ alogx = logx^a)

(∵ loga + logb = logab)

⇒

Applying exponential on both sides, we get,

⇒

Since y = vx, we get,

⇒

Cross multiplying on both sides we get,

⇒ x = C(x² – y²)

∴ The solution for the given Differential equation is x = C(X²-y²)

Question 8.

Solve the following equations:

Answer:

Given Differential equation is:

⇒

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that

⇒

Bringing like variables on same side we get,

⇒

We know that:

and

Also,

Integrating on both sides, we get,

⇒

(∵ log C is an arbitrary constant)

⇒

(∵ alogx = logx^a)

⇒

(∵ loga + logb = logab)

Since y = vx,

we get,

⇒

Applying exponential on both sides we get,

⇒

∴ The solution of the Differential equation is

Question 9.

Solve the following equations:

Answer:

Given Differential equation is:

⇒

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is y = vx.

Let us substitute this in (1)

⇒

We know that

⇒

Bringing like on the same side we get,

⇒

We know that

Integrating on both sides we get,

⇒

(∵ logC is an arbitrary constant)

⇒ log(1-2v²) = -4logx + 4logC

⇒ log(1–2v²) = -logx⁴ + logC⁴

(∵ xloga = loga^x)

⇒

(∵ )

Applying exponential on both sides we get,

⇒

Since y = vx, we get,

⇒

Cross multiplying on both sides we get,

⇒ x²(x²–2y²) = c⁴

⇒ x⁴–2x²y² = c⁴

∴ The solution for the given differential equation is x⁴–2x²y² = C⁴.

Question 10.

Solve the following equations:

Answer:

Given Differential equation is:

⇒

⇒ ……(1)

Homogeneous equation: A equation is said to be homogeneous if f(zx,zy) = zⁿf(x,y) (where n is the order of the homogeneous equation).

Let us assume:

⇒

⇒ f(zx,zy) = z⁰f(x,y)

So, given differential equation is a homogeneous differential equation.

We need a substitution to solve this type of linear equation and the substitution is x = vy.

Let us substitute this in (1)

⇒

We know that:

⇒

Bringing like variables on the same side we get,

⇒

We know that ∫e^xdx = e^x + C and

Integrating on both sides, we get,

⇒

⇒ e^v = logy + C

Since x = vy, we get

⇒

∴ The solution for the given Differential equation is .

Question 11.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

Question 12.

Solve the following differential equations :

Answer:

Here, (y² – 2xy)dx = (x² – 2xy)dy

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

x(y² – xy) = c

Question 13.

Solve the following differential equations :

Answer:

Here, 2xydx + ( x² + 2y² ) dy = 0

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

…… (1)

1 + 2v² = A – 2Av² + Bv² + cv

1 + 2v² = v²( – 2A + B) + cv + A

Comparing the coefficients of like power of v,

A = 1

C = 0

– 2A + B = 2

– 2 + B = 2

B = 4

Question 14.

Solve the following differential equations :

Answer:

Here, 3x²dy = (3xy + y²)dx

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 15.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 16.

Solve the following differential equations :

Answer:

Here, (x + 2y)dx – (2x – y)dy = 0

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 17.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

y + √(y² – x² ) = c

Question 18.

Solve the following differential equations :

Answer:

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

log log v = log|x| + logc

log v = xc

log y/x = xc

y = xe^xc

Question 19.

Solve the following differential equations :

Answer:

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 20.

Solve the following differential equations :

Answer:

y² + (x² – xy + y²)dy = 0

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 21.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Let 1 + v² = t

Differentiating both sides we get,

2vdv = dt

Question 22.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both sides we get,

tanv = – log|x| + logc

Question 23.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 24.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put x = vy

And

So,

Integrating both sides we get,

Itegration it by parts

Question 25.

Solve the following differential equations :

.

Answer:

Here,

It is a homogeneous equation

Put x = vy

And

So,

Integrating both sides wee get,

Question 26.

Solve the following differential equations :

Answer:

Here,

It is homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

…… (A)

Comparing the coefficient of like power of v

– A + C = 1 ……(i)

2A – B = 0

B = 2A ……(ii)

2B + 4C = 1 …… (iii)

Solving eq. (i),(ii) and (iii),

A = – 3/8,B = – 3/4, C = 5/8

Using eq.(A)

Question 27.

Solve the following differential equations :

Answer:

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

Question 28.

Solve the following differential equations :

Answer:

Here, x

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

tan v = – log x + log c

tan

Question 29.

Solve the following differential equations :

x

Answer:

Here, x

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

log (v + 2 log x + log c

log (v + log c

(v + c

Question 30.

Solve the following differential equations :

Answer:

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both sides we get,

log v – log sec v = – 2 log x + c

log () = log

Question 31.

Solve the following differential equations :

(x² + 3xy + y²)dx – x²dy=0

Answer:

Here, ()dx –

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

Question 32.

Solve the following differential equations :

Answer:

Here,

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

Question 33.

Solve the following differential equations :

Answer:

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides We get,

Comparing the coefficient of like power of v

A = – 3

C = 0

And 2A + B = – 1

⇒B = 5

So,

– 3 log v +

– 12 log v +

Question 34.

Solve the following differential equations :

Answer:

It is a homogeneous equation

Put y = vx

And

So,

Integrating Both Sides we get,

log (cosec v + cot v) = – log

log (cosec v + cot v) = log

Question 35.

Solve the following differential equations :

Answer:

It is a homogeneous equation,

Put y = vx

And

So,

Integrating both sides we get,

Let log v – 1 = t

t – log t = log

log v – log(log v – 1) = log

Question 36.

Solve each of the following initial value problems

(x² + y²)d x = 2xy dy, y(1) = 0

Answer:

()dx = 2xy dy, y(1) = 0

It is a homogenous equation

Put y = vx

And

So,

Integrating both sides we get,

log (1 – ) = – log(x) + log c

log (1 – ) = log

put v =

Put y = 0,x = 1 in eq. (1),

1 – 0 = c

c = 1

put value of c in eq,(1),

Question 37.

Solve each of the following initial value problems

,y(e) = 0

Answer:

,y(e) = 0

It is a homogeneous equation

Put y = vx

And

So,

On integrating both sides,

log xc

v = log(log xc)

put value of v,

log(log x) + k ……(1)

Put y = 0, x = e

0 = e log(log e) + k

k = 0

put in eq. (1),

y = x log(log(x))

Question 38.

Solve each of the following initial value problems

,y(1) = 0

Answer:

,y(1) = 0

It is a homogeneous equation

Put y = vx

And

So,

Integrating both sides we get,

– cos v = – log x + c

put value of v,

– cos = – log x + c …… (1)

Put y = 0, x = 1,We have

c = – 1

Now,

– cos = – log x – 1

log x = cos – 1

Question 39.

Solve each of the following initial value problems

(xy–y²)dx+x²dy=0,y(1) = 1

Answer:

(xy)dxdy = 0,y(1) = 1

It is a homogeneous equation

Put y = vx

And

So,

Put y = 1, x = 1

c = 1

Using equation(1),

x = y(log x + 1)

Question 40.

Solve each of the following initial value problems

,y(1) = 2

Answer:

,y(1) = 2

It is a homogeneous equation

Put y = vx

And

So,

…… (1)

2 + v = (A + B)v –A

Comparing the coefficient of like power of v,

A = – 2

A + B = 1

– 2 + B = 1

B = 3

Using in eq. (1)

– 2 log v + 3log (v – 1) = log x + c

Put the value of v,

Question 41.

Solve each of the following initial value problems

,y(1) = 1

Answer:

Here, y(1) = 1

It is homogeneous equation

Put y = vx

And

So,

log () = log

() = …… (1)

Put value of v,

() = c

Put y = 1,x = 1

c = 8

put in eq. (1),

() = 8

Question 42.

Solve each of the following initial value problems

Answer:

…… (1)

Put v =

And

So,

Integrating both sides,

– cot v = log x + c

– cot () = log x + c …… (2)

Put x = 1, y = in eq. (2)

c = – 1

put in eq.(2),

– cot () = log x – 1

Question 43.

Solve each of the following initial value problems

,y(2) = x

Answer:

,y(2) = x

It is homogeneous equation,

Put y = vx

And

So,

cosec v dv =

– log (cosce v + cot v) = – log x + c

Put y = ,x = 2,We have,

c = 0.301

now,

– log (cosec) = – log x + 0.301

Question 44.

Find the particular solution of the differential equation given that when x = 1, y = π/4

Answer:

Consider the given equation

this is homogeneous equation,

Put y = vx

And

So,

cos v dv =

sin v = log x + c

…… (1)

Put x = 1, y = in eq.(1),

c =

now,

Question 45.

Find the particular solution of the differential equation given that when x = 1, y = 0

Answer:

Consider the given equation,

It is a homogeneous equation

Put y = vx

And

So,

…… (1)

Put x = 1, y = 0 in eq.(1)

c =

Thus,

Question 46.

Find the particular solution of the differential equation given that when y = 1, x = 0

Answer:

…… (1)

Let v =

From (1) we have,

Integrating on both sides we have

…… (2)

Put x = 0,y = 1

0 = log(1) + c

c = 0

From equation (2) we have

Question 47.

Show that the family of curves for which is given by

Answer:

Here,

C = …… (i)

Differentiate both side,

2x dx – C dx = 2y dy

…… (ii)

Put equation (i) in equation(ii),We get,

Hence prove.

Differential Equations

Class 12th Mathematics RD Sharma Volume 2 Solution

Exercise 22.1

Exercise 22.10

Exercise 22.11

Very Short Answer

Mcq

Exercise 22.2

Exercise 22.3

Exercise 22.4

Exercise 22.5

Exercise 22.6

Exercise 22.7

Exercise 22.8

Exercise 22.9

Class 12^th Mathematics RD Sharma Volume 2 Solution