Differential Equations

It is given that equation is

⇒ y”” + sin(y’’’) = 0

We can see that the highest order derivative present in the differential is y””.

Thus, its order is four. The given differential equation is not a polynomial equation in its derivative.

Therefore, its degree is not defined.

It is given that equation is y’ + 5y = 0

We can see that the highest order derivative present in the differential is y’.

Thus, its order is one. It is polynomial equation in y’. The highest power raised to y’ is 1.

Therefore, its degree is one.

It is given that equation is

We can see that the highest order derivative present in the given differential equation is

Thus, its order is two. It is polynomial equation in and

Therefore, its degree is one.

It is given that equation is

We can see that the highest order derivative present in the given differential equation is.

Thus, its order is two. The given differential equation is not a polynomial equation in its derivative.

Therefore, its degree is not defined.

It is given that equation is

We can see that the highest order derivative present in the given differential equation is

Thus, its order is two. It is polynomial equation in and the power is 1.

Therefore, its degree is one.

It is given that equation is (y”’)² + (y”)³ +(y’)⁴ + y⁵ = 0

We can see that the highest order derivative present in the differential is y”’.

Thus, its order is three. It is polynomial equation in y”’, y” and y’

So, the highest power raised to y”’ is 2.

Therefore, its degree is two.

It is given that equation is y″′ +2y” + y’ = 0

We can see that the highest order derivative present in the differential is y”’.

Thus, its order is three. It is polynomial equation in y”’, y” and y’

So, the highest power raised to y”’ is 1.

Therefore, its degree is one.

It is given that equation is y’ + y = e^x

⇒ y’ + y – e^x = 0

We can see that the highest order derivative present in the differential is y’.

Thus, its order is one. It is polynomial equation in y’

So, the highest power raised to y’ is 1.

Therefore, its degree is one.

It is given that equation is y’’ + (y’)² + 2y = 0

We can see that the highest order derivative present in the differential is y”.

Thus, its order is two. It is polynomial equation in y” and y’

So, the highest power raised to y” is 1.

Therefore, its degree is one.

It is given that equation is y’’ + 2y’ + siny = 0

We can see that the highest order derivative present in the differential is y”.

Thus, its order is two. It is polynomial equation in y” and y’

So, the highest power raised to y” is 1.

Therefore, its degree is one.

It is given that equation is

We can see that the highest order derivative present in the given differential equation is.

Thus, its order is three.

The given differential equation is not a polynomial equation in its derivative.

Therefore, its degree is not defined.

It is given that equation is

We can see that the highest order derivative present in the given differential equation is .

Thus, its order is two.

It is given that y = e^x + 1

Now, differentiating both sides w.r.t. x, we get,

Now, Again, differentiating both sides w.r.t. x, we get,

⇒ y” = e^x

Now, Substituting the values of y’ and y” in the given differential equations, we get,

y” – y’ = e^x - e^x = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y = x² + 2x + C

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = 2x + 2

Now, Substituting the values of y’ in the given differential equations, we get,

y’ – 2x -2 = 2x + 2 – 2x - 2 = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y = cosx + C

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = -sinx

Now, Substituting the values of y’ in the given differential equations, we get,

y’ + sinx = -sinx + sinx = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y =

Now, differentiating both sides w.r.t. x, we get,

Therefore, LHS = RHS

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y = Ax

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = A

Now, Substituting the values of y’ in the given differential equations, we get,

xy’ = x.A = Ax = y = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y = xsinx

Now, differentiating both sides w.r.t. x, we get,

⇒ y’ = sinx + xcosx

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = xy’ = x(sinx + xcosx)

= xsinx + x²cosx

= y +

= RHS

Therefore, the given function is the solution of the corresponding differential equation.

It is given that xy = log y + C

Now, differentiating both sides w.r.t. x, we get,

⇒ y + xy’ =

⇒ y² + xyy’ = y’

⇒ (xy – 1)y’ = -y²

⇒ y’ =

Thus, LHS = RHS

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y – cosy = x

Now, differentiating both sides w.r.t. x, we get,

⇒ y’(1 + siny) = 1

⇒ y’ =

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = (ysiny + cosy + x)y’

= y = RHS

Therefore, the given function is the solution of the corresponding differential equation.

It is given that x + y = tan^-1y

Now, differentiating both sides w.r.t. x, we get,

Now, Substituting the values of y’ in the given differential equations, we get,

LHS = y²y’ + y² + 1 =

= -1 – y² + y² + 1

= 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

It is given that

Now, differentiating both sides w.r.t. x, we get,

Now, Substituting the values of y’ in the given differential equations, we get,

LHS =

= x –x = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

As we know that the number of constant in the general solution of a differential equation of order n is equal to its order.

Thus, the number of constant in the general equation of fourth order differential equation is four.

In a particular solution of a differential equation, there is no arbitrary constant.

The given equation is

Now, differentiating both sides w.r.t x, we get,

Now, again differentiating both sides, we get,

⇒ y’’ = 0

Therefore, the required differential equation is y’’ = 0.

The given equation is y² = a(b² – x²)

Now, differentiating both sides w.r.t x, we get,

⇒ 2yy’ = -2ax

⇒ yy’ = -ax -------(1)

Now, again differentiating both sides, we get,

y’.y’ +yy’’ = -a

⇒ (y’)² + yy” = -a --------(2)

Now, dividing equation (2) by (1), we get,

⇒ xyy” + x(y’)² – yy” = 0

Therefore, the required differential equation is xyy” + x(y’)² – yy” = 0.

It is given that y = ae^3x + be^-2x --------(1)

Now, differentiating both side we get,

y’ = 3ae^3x - 2be^-2x --------(2)

Now, again differentiating both sides, we get,

y’’ = 9ae^3x + 4be^-2x -------(3)

Now, let us multiply equation (1) with 2 and then adding it to equation (2), we get,

(2ae^3x + 2be^-2x) + (3ae^3x - 2be^-2x) = 2y – y’

⇒ 5ae^3x = 2y + y’

Now, let us multiply equation (1) with 3 and subtracting equation (2), we get

(3ae^3x + 3be^-2x) - (3ae^3x - 2be^-2x) = 3y – y’

⇒ 5be^-2x = 3y - y’

y” = 9. +4

⇒ y” = 6y + y’

⇒ y” – y’ - 6y = 0

Therefore, the required differential equation is y” – y’ - 6y = 0.

It is given y = e^2x(a + bx) -------(1)

Now, differentiating both side w.r.t. x, we get,

y’ = 2e^2x(a + bx) + e^2x.b ------(2)

Now, let us multiply equation (1) with 2 and then subtracting it to equation (2), we get,

y’ – 2y = e^2x(2a +2bx + b) – e^2x(2a + 2bx)

⇒ y’ – 2y = be^2x ---------(3)

Now, again differentiating both sides w.r.t. x, we get,

y” – 2y’ = 2be^2x ------(4)

Dividing equation (4) by equation (3), we get,

⇒ y” – 2y’ = 2y’ – 4y

⇒ y” – 4y’ – 4y = 0

Therefore, the required differential equation is y” – 4y’ - 4y = 0.

It is given that y = e^x(acosx + bsinx) ------(1)

Now, differentiating both w.r.t. x, we get,

y’ = e^x(acosx + bsinx) + e^x(-asinx + bcosx)

⇒ y’ = e^x[(a + b)cosx – (a – b)sinx)] ------(2)

Again, differentiating both sides w.r.t. x, we get,

y” = e^x[(a + b)cosx – (a – b)sinx)] + e^x[-(a + b)sinx – (a – b)cosx)]

⇒ y” = e^x[2bcosx – 2asinx]

⇒ y” = 2e^x(bcosx – asinx) ----(3)

Adding equation (1) and (3), we get,

⇒ 2y + y” = 2y’

Therefore, the required differential equation is 2y + y” = 2y’= 0.

The center of the circle touching the y- axis at orgin lies on the x – axis.

Let (a,0) be the centre of the circle.

Thus, it touches the y – axis at orgin, its radius is a.

Now, the equation of the circle with centre (a,0) and radius (a) is

(x –a)² – y² = a²

⇒ x² + y² = 2ax

Now, differentiating both sides w.r.t. x , we get,

2x + 2yy’ = 2a

⇒ x + yy’ = a

Now, on substituting the value of a in the equation, we get,

x² + y² = 2(x + yy’)x

⇒ 2xyy’ + x² = y²

Therefore, the required differential equation is 2xyy’ + x² = y² .

We know that the equation of the parabola having the vertex at origin and the axis along the positive y- axis is

x² = 4ay ------(1)

Now, differentiating equation (1) w.r.t. x, we get,

2x = 4ay’ -----(2)

On dividing equation (2) by equation (1), we get,

⇒ xy’ = 2y

⇒ xy’ – 2y = 0

Therefore, the required differential equation is xy’ – 2y = 0.

We know that the equation of the family of ellipses having foci on y – axis and the centre at origin is

-------(1)

Now, differentiating equation (1) w.r.t. x, we get,

----(2)

Now, again differentiating w.r.t. x, we get,

Let us substitute the value in eq. (2), we get,

⇒ -x(y’)² – xyy” + yy’ = 0

⇒ xyy” + x(y’)² – yy’ = 0

Therefore, the required differential equation is xyy” + x(y’)² – yy’ = 0.

We know that the equation of the family of hyperbolas having foci on x – axis and the centre at origin is

-------(1)

Now, differentiating equation (1) w.r.t. x, we get,

--------(2)

Now, again differentiating w.r.t. x, we get,

Let us substitute the value in eq. (2), we get,

⇒ x(y’)² + xyy” - yy’ = 0

⇒ xyy” + x(y’)² – yy’ = 0

Therefore, the required differential equation is xyy” + x(y’)² – yy’ = 0.

Let the centre of the circle on y – axis be (0,b).

We know that the differential equation of the family of circles with centre at (0, b) and radius 3 is: x² + (y- b)² = 3²

⇒ x² + (y- b)² = 9 ----(1)

Now, differentiating both sides w.r.t. x, we get,

2x + 2(y – b).y’ = 0

⇒ (y – b). y’ = x

⇒ y – b =

Thus, substituting the value of ( y – b) in equation (1), we get,

⇒ x²((y’)² + 1) = 9(y’)²

⇒ (x² – 9)(y’)² + x² = 0

Therefore, the required differential equation is (x² – 9)(y’)² + x² = 0

It is given that y = c₁e^x + c₂e^-x

Now, differentiating both sides w.r.t. x, we get,

Again, differentiating both sides w.r.t. x, we get,

Therefore, the required differential equation is .

It is given that that y = x

Now, differentiating both sides w.r.t. x, we get,

Again, differentiating both sides w.r.t. x, we get,

Now, substitute the value of in the given options, we will see that the differential equation in option (C) is correct.

= -x² + x²

= 0

⇒

dy = (1 - y) dx

Separating variables

⇒ -log(1-y) = x + logc

⇒ -log(1-y) - logc = x

⇒ log (1-y)c = -x

Or

⟹

Dividing both sides by (tanx)(tany)

Integrating both sides,

⇒ lettan x = t &tany = u

⇒ log t = -log u + log c

Or,

⇒ log(tanx) = -log(tany) + log c

Or

⇒ (tan x) (tan y) = c

Integrating both sides,

Let ( = t

Then,

So,

⇒ y = log t

Or,

+ c

Separating variables,

Integrating both sides,

y log y dx – x dy = 0

⇒ (y log y) dx = xdy

Separating variables,

Integrating both sides,

⇒ let logy = t

⇒ log x + log c = log t

Or,

⇒ log x + log c = log (log y)

⇒ log cx = log y

Or,

⇒ log y = cx

Separating variables,

Or,

Integrating both sides,

Letbe a constant,

Or,

0r,

Separating variables,

Integrating both sides,

Now to integrate we have to multiply it by 1
because,

∵ { - i)

So,

According to i)

Let u be and v be 1

We can take the values of u and v from the formula (I.L.A.T.E)

So,

Or

putting the value of t,

Separating the variables,

Now Integrating both sides,

Let &

Then,

Then,

Or,

⇒ log t = log u + log c

Substituting the values of t and u on above equation.

Or,

(

Separating variables,

Integrating both sides,

Integrating it partially,

Now comparing the coefficients of

⇒ A + B = 2

⇒ B + C = 1

⇒ A + C = 0

Solving them we will get the values of A,B,C

Putting the values of A,B,C in i)

So,

Then, 2xdx = dt

or,

Or,

Now, we are given that y = 1 when x = 0

So,

C = 1

Putting the value of c in ii)

Separating variables,

Or,

Integrating both sides,

Now let,

Or

Now comparing the values of A,B,C

A + B + C = 0

B-C = 0

A = -1

Solving these we will get that

Now putting the values of A,B,C in ii)

Now integrating it,

Now we are given that

Or,

Now putting the value of

Then,

Separating variables,

Now y = 2 when x = 0

⇒ 2 = 0 + c

⇒ c = 2

Putting the value of ini)

Separating variables,

Integrating both sides,

⇒ log y = -log (cos x) + log c

Or,

⇒ log y = log (sec x) + log c

⇒ log y = log c (sec x)

⇒ y = c (sec x) -i)

Now we are given that y = 1 when x = 0

⇒ 1 = c (sec 0)

⇒ 1 = c × 1

⇒ c = 1

Putting the value of c in i)

⇒ y = sec x

To find the equation of a curve that passes through point(0,0) and has differential equation

So, we need to find the general solution of the given differential equation and the put the given point in to find the value of constant.

Separating variables,

Integrating both sides,

{Using the formula, )

Now let

integrating for .

Or,

Or,

Now we are given that the curve passes through point(0,0)

Putting the value of C in ii)

So,

So the required equations become,

For this question, we need to find the particular solution at point(1,-1) for the given differential equation.

Given differential equation is

Separating variables,

Or,

Integrating both sides,

Now separating like terms on each side,

Or,

Now we are given that, the curve passes through (1, -1)

⇒ -2-c = log (1)

⇒ c = -2 + 0 (∵ log(1) = 0)

So c = -2

Putting the value of c in

We know that slope of a tangent is =

So we are given that the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Now separating variables,

⇒ ydy = xdx

Integrating both sides,

Now the curve passes through (0, -2).

∴ 4-0 = 2c

⇒ c = 2

Putting the value of c in i)

_{We know that (x,y) is the point of contact of curve and its tangent.}

slope(m1)for line joining (x,y) and (-4,-3) is

Also we know that slope of tangent of a curve is .

Now, according to the question,

(m2) = 2(m1)

Separating variables,

Integrating both sides,

⇒ log(y + 3) = 2log(x + 4) + log c

Now, this equation passes thorough the point (-2,1).

⇒ 4 = 4c

⇒ c = 1

Substitute the value of c in iii)

Let the rate of change of the volume of the balloon be k. (k is a constant)

Or,

Integrating both sides,

Now, given that

At t = 0, r = 3:

⇒ 4π × 33 = 3(k×0 + c)

⇒ 108π = 3c

⇒ c = 36π

At t = 3, r = 6:

⇒ k = 84π

Substituting the values of k and c in i)

So the radius of balloon after t seconds is

let t be time

p be principal

r be rate of interest

according the information principal increases at the rate of r% per year.

Separating variables,

Integrating both sides,

Given that t = 0, p = 100.

Now, if t = 10, then p = 2×100 = 200

So,

So r is 6.93%.

Let p and t be principal and time respectively.

Given that principal increases continuously at rate of 5% per year.

Separating variables,

Integrating both sides,

When t = 0, p = 1000

At t = 10

So after 10 years the total amount would be Rs.1648

let y be the number of bacteria at any instant t.

Given that the rate of growth of bacteria is proportional to the number present

Separating variables,

Integrating both sides,

⇒ log y = kt + c -i)

Let y’ be the number of bacteria at t = 0.

⇒ log y’ = c

Substituting the value of c in

⇒ log y = kt + log y’

⇒ log y- log y’ = kt

Also, given that number of bacteria increases by 10% in 2 hours.

Substituting this value in

So, becomes

Now, let the time when number of bacteria increase from 100000 to 200000 be t’.

So from

So bacteria increases from 100000 to 200000 in hours.

separating variables

Integrating both sides

Or,

(c is a constant)

So the correct option is A.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

(x² + xy)dy = (x² + y²)dx

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating on both side,

- v - 2log|1 - v| = log|x| + logc

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

v = logx + C

y = xlogx + Cx

The required solution of the differential equation.

(x - y)dy = (x + y)dx

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

(x - y)dy – (x + y)dx = 0

To make it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides we get,

2vdv = dt

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

Put 1 + v² = t

2vdv = dt

log(t)

∴ log(1 + v²) = -logx + logC (∴ From (i) eq.)

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

log secv – logv = 2logkx

The required solution of the differential equation.

Here, putting x= kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both side, we get

log(cosecv – cotv) = -logx + logC

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

Put, logv – 1 = t

logt

log(logv - 1)

∴ log(logv - 1) – log(v) = log(x) + log(c) (From (i) eq.)

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

x = vy

Differentiation eq. with respect to x, we get

Integrating both sides, we get

Put e^v + v = t

(e^v + 1)dv = dt

logt

log(e^v + v)

∴ log(e^v + v) = - logy + logC (∴ From (i) eq.)

Multiply by y on both side, we get

ye^x/y + x = C

x + ye^x/y = C

The required solution of the differential equation.

(x + y)dy + (x - y)dx = 0

Here, putting x= kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

(x + y)dy + (x - y)dx = 0

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 1 when x = 1

The required solution of the differential equation.

x²dy + (xy + y²)dx = 0

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

x²dy + (xy + y²)dx = 0

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 1 when x = 1

3x²y = y + 2x

y + 2x = 3x²y

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

∫cosec²vdv = - logx – logC

-cot v = - logx – logC

cot v = logx + logC

1 = C

e¹ = C

The required solution of the differential equation.

Here, putting x= kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

- cosv = - logx + C

y = 0 when x = 1

- 1 = C

The required solution of the differential equation.

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 2 when x = 1

- 1 = C

The required solution of the differential equation.

Therefore, we shall substitute,

x = vy

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

It cannot be homogeneous as we can see that (4x + 6y + 5) is not homogeneous.

(B) (xy)dx – (x³ + y³)dy = 0

It cannot be homogeneous as the xy which multiplies with dx and x³ + y³ which multiplies with dy are not of same degree.

(C) (x³ + 2y²)dx + 2xy dy = 0

Similarly, it cannot be homogeneous as the x³ + 2y² which multiplies with dx and 2xy which multiplies with dy are not of same degree.

(D) y²dx + (x² – xy – y²)dy = 0

Here, putting x = kx and y = ky

= k⁰.f(x,y)

Therefore, the given differential equations is homogeneous.

It is given that

This is equation in the form of (where, p = 2 and Q = sinx)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

--------(1)

Let I =

Now, putting the value of I in (1), we get,

Therefore, the required general solution of the given differential equation is

It is given that

This is equation in the form of (where, p = 3 and Q = )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

⇒ ye^3x = e^x + C

⇒ y = e^-2x + Ce^-3x

Therefore, the required general solution of the given differential equation is y = e^-2x + Ce^-3x

It is given that

This is equation in the form of (where, p = and Q = )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

Therefore, the required general solution of the given differential equation is .

It is given that

This is equation in the form of (where, p = secx and Q = tanx)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

⇒ y(secx + tanx) = secx + tanx – x+ C

Therefore, the required general solution of the given differential equation is

y(secx + tanx) = secx + tanx – x+ C.

It is given that

This is equation in the form of (where, p = and Q =)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

-----------(1)

Now, Let t = tanx

⇒ sec²xdx = dt

Thus, the equation (1) becomes,

⇒ te^tanx = (t – 1)e^t + C

⇒ te^tanx = (tanx – 1)e^tanx + C

⇒ y = (tanx -1) + C e^-tanx

Therefore, the required general solution of the given differential equation is

y = (tanx -1) + C e^-tanx.

It is given that

This is equation in the form of (where, p = and Q =xlogx)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

Therefore, the required general solution of the given differential equation

It is given that

This is equation in the form of (where, p = and Q =)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

------------------(1)

Now,

Now, substituting the value in (1), we get,

Therefore, the required general solution of the given differential equation is

It is given that

This is equation in the form of (where, p = and Q = )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

Therefore, the required general solution of the given differential equation is

It is given that

This is equation in the form of (where, p = and Q = 1 )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) =

⇒ y(xsinx) = -xcosx + sinx + C

⇒

⇒

Therefore, the required general solution of the given differential equation is

.

It is given that

This is equation in the form of (where, p = -1 and Q = y )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) =

=> x = - y – 1 + Ce^y

=> x + y + 1 = Ce^y

Therefore, the required general solution of the given differential equation is

x + y + 1 = Ce^y.

It is given that

This is equation in the form of (where, p = and Q = y )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) =

Therefore, the required general solution of the given differential equation is .

It is given that

This is equation in the form of (where, p = and Q = 3y )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

x(I.F.) =

⇒ x = 3y² + Cy

Therefore, the required general solution of the given differential equation is x = 3y² + Cy.

It is given that

This is equation in the form of (where, p = 2tanx and Q =sinx )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

----------------(1)

Now, it is given that y = 0 at x =

⇒ 0 = 2 + C

⇒ C = -2

Now, Substituting the value of C = -2 in (1), we get,

⇒ y = cosx – 2cos²x

Therefore, the required general solution of the given differential equation is

y = cosx – 2cos²x.

It is given that

This is equation in the form of (where, p = and Q = )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

----------------(1)

Now, it is given that y = 0 at x = 1

0 = tan^-1 1+ C

⇒ C =

Now, Substituting the value of C = in (1), we get,

Therefore, the required general solution of the given differential equation is

It is given that

This is equation in the form of (where, p = -3cotx and Q = sin2x)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

⇒ y cosec³x = 2cosecx + C

⇒ y =

⇒ y = -2sin²x + Csin³x--------------(1)

Now, it is given that y = 2 when x =

Thus, we get,

2 = -2 + C

⇒ C = 4

Now, Substituting the value of C = 4 in (1), we get,

y = -2sin²x + 4sin³x

⇒ y = 4sin³x - 2sin²x

Therefore, the required general solution of the given differential equation is

y = 4sin³x - 2sin²x.

Let F(x,y) be the curve passing through origin and let (x,y) be a point on the curve.

We know the slope of the tangent to the curve at (x,y) is .

According to the given conditions, we get,

This is equation in the form of (where, p = -1 and Q = x )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

--------(1)

Now,

Thus, from equation (1), we get,

⇒ y = -(x+1) + Ce^x

⇒ x + y + 1 = Ce^x -----(2)

Now, it is given that curve passes through origin.

Thus, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get,

x + y – 1 = e^x

Therefore, the required general solution of the given differential equation is

x + y -1 = e^x

Let F(x,y) be the curve and let (x,y) be a point on the curve.

We know the slope of the tangent to the curve at (x,y) is .

According to the given conditions, we get,

This is equation in the form of (where, p = -1 and Q = x - 5)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

------------(1)

Now,

Thus, from equation (1), we get,

⇒ y = 4 – x + Ce^x

⇒ x + y – 4 = Ce^x

Now, it is given that curve passes through (0,2).

Thus, equation (2) becomes:

0 + 2 – 4 = C e⁰

⇒ - 2 = C

⇒ C = -2

Substituting C = -2 in equation (2), we get,

x + y – 4 =-2e^x

⇒ y = 4 – x – 2e^x

Therefore, the required general solution of the given differential equation is

y = 4 – x – 2e^x

It is given that

This is equation in the form of (where, p = and Q =2x)

Now, I.F. =

It is given that

This is equation in the form of (where, p = and Q = )

Now, I.F. =

It is given that equation is

We can see that the highest order derivative present in the differential is .

Thus, its order is two. It is polynomial equation in . The highest power raised to is 1.

Therefore, its degree is one.

It is given that equation is

We can see that the highest order derivative present in the differential is.

Thus, its order is one. It is polynomial equation in . The highest power raised to is 3.

Therefore, its degree is three.

It is given that equation is

We can see that the highest order derivative present in the differential is .

Thus, its order is four. The given differential equation is not a polynomial equation.

Therefore, its degree is not defined.

It is given that xy = a e^x + b e^–x + x²

Now, differentiating both sides w.r.t. x, we get,

Now, Again differentiating both sides w.r.t. x, we get,

Now, Substituting the values of ’ and in the given differential equations, we get,

LHS =

= x(ae^x +be^-x + 2) + 2(ae^x - be^-x + 2) –x(ae^x +be^-x + x²) + x² – 2

= (axe^x +bxe^-x + 2x) + 2(ae^x - be^-x + 2) –x(ae^x +be^-x + x²) + x² – 2

= 2ae^x -2be^-x +x² + 6x -2

≠ 0

⇒ LHS ≠ RHS.

Therefore, the given function is not the solution of the corresponding differential equation.

It is given that y = e^x(acosx + bsinx) = ae^xcosx + be^xsinx

Now, differentiating both sides w.r.t. x, we get,

Now, again differentiating both sides w.r.t. x, we get,

Now, Substituting the values of ’ and in the given differential equations, we get,

LHS =

=2e^x(bcosx – asinx) -2e^x[(a + b)cosx + (b –a ) sinx] + 2e^x(acosx + bsinx)

=e^x[(2bcosx – 2asinx) - (2acosx + 2bcosx) - (2bsinx – 2asinx) + (2acosx + 2bsinx)]

= e^x[(2b – 2a – 2b + 2a)cosx] + e^x[(-2a – 2b + 2a + 2bsinx]

= 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

It is given that y = xsin3x

Now, differentiating both sides w.r.t. x , we get,

Now, again differentiating both sides w.r.t. x, we get,

Now, substituting the value of in the LHS of the given differential equation, we get,

= (6.cos3x – 9xsin3x) + 9xsin3x – 6cos3x

= 0 = RHS

Therefore, the given function is the solution of the corresponding differential equation.

It is given that x² = 2y² log y

Now, differentiating both sides w.r.t. x, we get,

2x = 2.

Now, substituting the value of in the LHS of the given differential equation, we get,

= xy –xy

= 0

Therefore, the given function is the solution of the corresponding differential equation.

It is given that (x – a)² + 2y² = a²

⇒ x2 + a2 – 2ax + 2y2 = a2

⇒ 2y2 = 2ax – x2 ---------(1)

Now, differentiating both sides w.r.t. x, we get,

- ---------(2)

So, equation (1), we get,

2ax = 2y² + x²

On substituting this value in equation (3), we get,

Therefore, the differential equation of the family of curves is given as .

It is given that (x³–3xy²) dx = (y³–3x²y)dy

- --------(1)

Now, let us take y = vx

Now, substituting the values of y and in equation (1), we get,

On integrating both sides we get,

--------(2)

Now,

---------(3)

Let

⇒

⇒

⇒

Now,

Let v² = p

Now, substituting the values of I₁ and I₂ in equation (3), we get,

Thus, equation (2), becomes,

⇒ (x² – y²)² = C’⁴(x² + y² )⁴

⇒ (x² – y²) = C’²(x² + y² )

⇒ (x² – y²) = C(x² + y² ), where C = C’²

Therefore, the result is proved.

We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is :

(x -a)² + (y –a)² = a² -----------(1)

Now differentiating above equation w.r.t. x, we get,

2(x-a) + 2(y-a) = 0

⇒ (x – a) + (y – a)y’ = 0

⇒ x – a +yy’ – ay’ = 0

⇒ x + yy’ –a(1+y’) = 0

⇒ a =

Now, substituting the value of a in equation (1), we get,

⇒ (x - y)².y’² + (x – y)² = (x + yy’)²

⇒ (x – y)²[1 + (y’)²] = (x + yy’)²

Therefore, the required differential equation of the family of circles is

(x – y)²[1 + (y’)²] = (x + yy’)²

It is given that

On integrating, we get,

⇒ sin^-1x + sin^-1y = C

It is given that

On integrating both sides, we get,

Let

Then,

Now, let A = , then, we have,

It is given that sin x cos y dx + cos x sin y dy = 0

⇒ tanxdx + tanydy = 0

So, on integrating both sides, we get,

log(secx) + log(secy) = logC

⇒ log(secx.secy) = log C

⇒ secx.secy = C

The curve passes through point

Thus, 1× = C

⇒ C =

On substituting C = in equation (1), we get,

secx.secy =

Therefore, the required equation of the curve is

It is given that (1 + e^2x) dy + (1 + y²) e^x dx = 0

On integrating both sides, we get,

------(1)

Let e^x = t

⇒ e^2x = t²

⇒ e^xdx = dt

Substituting the value in equation (1), we get,

⇒ tan^-1 y + tan^-1 t = C

⇒ tan^-1 y + tan^-1 (e^x) = C -------(2)

Now, y =1 at x = 0

Therefore, equation (2) becomes:

tan^-1 1 + tan^-1 1 = C

Substituting in (2), we get,

tan^-1 y + tan^-1 (e^x) =

It is given that

------(1)

Let

Differentiating it w.r.t. y, we get,

------(2)

From equation (1) and equation (2), we get,

⇒ dz = dy

On integrating both sides, we get,

z = y + C

It is given that (x – y)(dx +dy) = dx – dy

⇒ (x –y + 1)dy = (1- x + y)dx

----------------(1)

Let x – y = t

Now, let us substitute the value of x-y and in equation (1), we get,

-------(2)

On integrating both side, we get,

t + log|t| = 2x + C

⇒ ( x – y ) + log |x – y| = 2x + C

⇒ log|x – y| = x + y + C --------(3)

Now, y = -1 at x = 0

Then, equation (3), we get,

log 1 = 0 -1 + C

⇒ C = 1

Substituting C = 1in equation (3), we get,

log|x – y| = x + y + 1

Therefore, a particular solution of the given differential equation is log|x – y| = x + y + 1.

It is given that

This is equation in the form of (where, p = and Q = )

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

It is given that

This is equation in the form of (where, p = cotx and Q = 4xcosecx)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

-----------(1)

Now, y = 0 at x =

Therefore, equation (1), we get,

0 =

⇒ C =

Now, substituting C = in equation (1), we get,

ysinx =

Therefore, the required particular solution of the given differential equation is

ysinx =

It is given that

On integrating both sides, we get,

----------------(1)

Let

⇒ e^ydt = -dt

Substituting value in equation (1), we get,

⇒ -log|t| = log|C(x+1)|

⇒ -log|2 – e^y| = log|C(x + 1)|

------------------(2)

Now, at x = 0 and y = 0, equation (2) becomes,

⇒ C = 1

Now, substituting the value of C I equation (2), we get,

Therefore, the required particular solution of the given differential equation is

Let the population at any instant (t) be y.

Now it is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

Now, integrating both sides, we get,

log y = kt + C ------(1)

According to given conditions,

In the year 1999, t = 0 and y = 20000

⇒ log20000 = C ----(2)

Also, in the year 2004, t = 5 and y = 25000

⇒ log 25000 = k.5 + C

⇒ log 25000 = 5k + log 20000

--------(3)

Also, in the year 2009, t = 10

Now, substituting the values of t, k and c in equation (1), we get

logy =

⇒ y = 31250

Therefore, the population of the village in 2009 will be 31250.

It is given that

Integrating both sides, we get,

log|x| - log|y| = log k

⇒

⇒

⇒

⇒ y = Cx where C =

The integrating factor of the given differential equation

is

Thus, the general solution of the differential equation is given by,

It is given that e^xdy + (ye^x + 2x) dx = 0

This is equation in the form of (where, p = 1 and Q = -2xe^-x)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

⇒ ye^x = -x² + C

⇒ ye^x + x² = C