Differential Equations

given:

To find: solution of the given differential equation.

Re-writing the equation as

Now integrating both sides

Formula:

(c is some arbitrary constant)

(d is some arbitrary constant = c ln2)

To find: differential equation of all non-vertical lines

General form of equation of line is given by the equation.

y=mx+c where m is the slope of the line

for the given condition the slope of line cannot be equal to or because if it so then the line will become perpendicular which is not required

so and

now differentiating the general form of equation of line

Formula:

Differentiating it again we get

Therefore, this is the differential equation of all non-vertical lines.

given: and (5,0) is a solution of this equation

To find: solution of the given differential equation

Re-writing the equation as

Now integrating both sides

Formula:

Now it is given that (5,0) is solution to this equation, so satisfying these values to find c

Hence the solution is

e^2y = 2x – 9

When y = 3 then,

e²⁽³⁾ = 2x – 9

e⁶ = 2x – 9

e⁶ + 9 = 2x

given:

To find: solution of the given differential equation

Re-writing the given equation as

It is a first order linear differential equation

Comparing it with

Calculating integrating factor

Formula:

Therefore, the solution of the differential equation is given by

Formula:

given:

To find: solution of the given differential equation

Formula:

Re-writing the given equation as

Now integrating both sides

ln y – ln c = x – x²

given:

To find: solution of the given differential equation

It is a first order linear differential equation

Comparing it with

p(x)=a

q(x)=e^xm

Calculating integrating factor

Therefore, the solution of the differential equation is given by

Formula:

given:

To find: solution of the given differential equation

Assuming x+y=t

Differentiating both sides with respect to x

Substituting in the above equation

Re-writing the equation as

Integrating both sides

Formula:

Substituting t = x + y

is the solution of the given differential equation

given: ydx – xdy =x²dx

To find: solution of the given differential equation

Re-writing the given equation as

Integrating both sides

Formula:

given: and (0,0) is a solution of the equation

To find: solution of the given differential equation

Re-writing the equation as

Now integrating both sides

Formula:

Substituting (0,0) to find the value of c

0=0+c

c=0

therefore, the solution is

given:

To find: solution of the given differential equation

Re-writing the equation as

It is a first order linear differential equation

Comparing it with

q(y)=2y²

Calculating integrating factor

Formula:

Formula:

Therefore, the solution of the differential equation is given by

Formula:

x³ = y³ + cy

given:

To find: solution of the given differential equation

Re-writing the equation as

Integrating both sides

Let sin x=t

cos x dx=dt

formula:

Formula:

ln(1+y) = -ln(2+t)+log c

ln(1+y)+ln(2+sin x )=log c

(1+y)(2+sinx) = c

When x = 0 and y = 1

c = 4

If x = π/2 then,

given: (0,-1) is a solution

To find: solution of the given differential equation

Re-writing the given equation as

It is a first order linear differential equation

Comparing it with

Calculating integrating factor

Formula:

Therefore, the solution of the differential equation is given by

(

Formula:

Substituting (0, -1) to find the value of c

-1=-1+c

c=0

is the solution

therefore y (1) is

given:

To find: differential equation for the given equation

Differentiating both sides

Formula:

Differentiating again on both sides

Formula:

Hence the solution is

To find: differential equation of all circles which pass through origin and centre lies on y axis

Assume a point (0, k) on the y axis

Therefore, the radius of circle is given as

And the general form of equation of circle is

(x-a)²+(y-b)²=r²

Where (a, b) is the centre and r is radius, now substituting the value in the above equation we get

(x-0)²+(y-k)²=k²

x²+y²-2yk=0 …. (i)

Differentiating both side with respect to x

Formula:

Substituting the value of k in eq(i)

given: and (0,0) is a solution to the curve

To find: equation of curve satisfying this differential equation

Re-writing the equation as

Comparing it with

Calculating integrating factor

Calculating

Assume 1+x²=t

2x dx=dt

Formula:

Substituting t=1+x²

IF=1+x²

Therefore, the solution of the differential equation is given by

Formula:

Satisfying (0,0) in the curve equation to find c

0=0+c

c=0

therefore, the equation of curve is

given:

To find: solution of the given differential equation

Re-writing the given equation as

It is clearly a homogenous differential equation

Assuming y=vx

Differentiating both sides

Substituting from the given equation

Now integrating both sides

Formula:

Substituting

is the solution of the given differential equation

given:

To find: solution of given differential equation

Re-writing the given equation as

It is a first order linear differential equation

Comparing it with

Calculating integrating factor

Formula:

Therefore, the solution of the differential equation is given by

Assume

Differentiating both sides

Formula:

Formula:

Substituting t

given:

To find: solution of given differential equation

Re-writing the given equation as

It is clearly a homogenous differential equation

Assuming x=vy

Differentiating both sides

Substituting in the given equation

Now substituting

Integrating both sides

Formula:

Substituting

given: (x+y) (dx – dy) =dx+dy

To find: solution of given differential equation

Re-writing the given equation as

Assume x+y=z

Differentiating both sides wrt to x

Substituting this value in the given equation

Now integrating both sides

Formula:

Substituting z=x+y

x-y-ln(x + y)-c=0

ln(x + y) + ln c = x – y

ln c(x + y) = x – y

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

x + y = 1/c (e^x-y)

x + y = d e^x-y

where d = 1/c

given: and (1, -2) is a solution

To find: solution of given differential equation

Re-writing the equation as

Integrating both sides

Formula:

Substituting (-2,1) to find the value of c

0=-2+c

c=2

⇒ 2 ln x=y-3 ln (y+3) +2

⇒ 2 ln x +3 ln (y+3) =y+2

⇒ 2 ln x +3 ln (y+3) =y+2

⇒ ln x² + ln (y+3)³ =y+2

⇒ x²(y+3)³ = y + c

given:

is a solution of the given equation

To find: solution of given differential equation

Re-writing the given equation as

It is a first order linear differential equation

Comparing it with

p(x)=cot x

q(x)=2 cos x

Calculating integrating factor

Formula:

IF = sin x

Therefore, the solution of the differential equation is given by

Substituting to get the value of c

Therefore, the solution is

given: Ax² + By²=1

To find: differential equation of the given curve

Differentiating the given curve with respect to x

(i)

Formula:

Again, differentiating the curve (i) we get,

Substituting this in eq(i)

given: (1+y²) tan^-1x dx + 2y(1+x²) dy=0

To find: solution of given differential equation

Re-writing the equation as

Integrating both sides

For LHS

Assume tan^-1 = t

Formula:

For RHS

Assume 1+y²=z

2y dy=dz

Now substituting and integrating both sides

Formula:

Substituting for t and z

Is the solution for the given differential equation

To find: differential equation of system of concentric circles with centre (1, 2)

General equation of such curve is given by

(x-a)²+(y-b)²=k² where (a, b) is the centre and k as radius

Now substituting the values

(x-1)²+(y-2)²=k²

Differentiating the curve with respect to x

Now means differentiation of xy with respect to x

Using product rule

Putting it back in original given differential equation

Divide by x

Compare with

we get and Q = sinx + logx

This is linear differential equation where P and Q are functions of x

For the solution of linear differential equation, we first need to find the integrating factor

⇒ IF = e^∫Pdx

⇒ IF = e^2logx

⇒ IF = x²

The solution of linear differential equation is given by y(IF) = ∫Q(IF)dx + c

Substituting values for Q and IF

⇒ yx² = ∫(sinx + logx)x²dx + c

⇒ yx² = ∫x²sinxdx + ∫x²logxdx + c …(a)

Let us find the integrals ∫x²sinxdx and ∫x²logxdx individually

Using uv rule for integration

⇒ ∫uvdx = u∫vdx - ∫(u’∫v)dx

⇒ ∫x²sinxdx = x²(-cosx) - ∫2x(-cosx)dx

⇒ ∫x²sinxdx = -x²cosx + 2∫xcosxdx

⇒ ∫x²sinxdx = -x²cosx + 2(xsinx - ∫sinxdx)

⇒ ∫x²sinxdx = -x²cosx + 2(xsinx – (-cosx))

⇒ ∫x²sinxdx = -x²cosx + 2xsinx +2cosx …(i)

Now ∫x²logxdx

Again, using product rule

Substitute (i) and (ii) in (a)

Divide by x²

Given: (1 + tany) (dx – dy) + 2xdy = 0

⇒ dx – dy + tany dx – tany dy + 2xdy = 0

Divide throughout by dy

Divide by (1 + tany)

Compare with

we get and Q = 1

This is linear differential equation where P and Q are functions of y

For the solution of linear differential equation, we first need to find the integrating factor

⇒ IF = e^∫Pdy

Put

Add and subtract siny in numerator

Consider the integral

Put siny + cosy = t hence differentiating with respect to y

we get which means

dt = (cosy – siny)dy

Resubstitute t

Hence the IF will be

⇒ IF = e^{y + log(siny+cosy)}

⇒ IF = e^y × e^{log(siny+cosy)}

⇒ IF = e^y(siny + cosy)

The solution of linear differential equation is given by x(IF) = ∫Q(IF)dy + c

Substituting values for Q and IF

⇒ xe^y(siny + cosy) = ∫(1)e^y(siny + cosy)dy + c

⇒ xe^y(siny + cosy) = ∫(e^ysiny + e^ycosy)dy + c

Put e^ysiny = t and differentiating with respect to y we get which means dt = (e^ysiny + e^ycosy)dy

Hence

⇒ xe^y(siny + cosy) = ∫dt + c

⇒ xe^y(siny + cosy) = t + c

Resubstituting t

⇒ xe^y(siny + cosy) = e^ysiny + c

Using the hint given and substituting x + y = z

Differentiating z – x with respect to x

Integrate

We know that cos 2z = 2 cos²z – 1

and sin 2z = 2 sin z cos z

Let hence differentiating with respect to z

we get

hence

⇒ log t + c = x

Resubstituting t

Resubstitute z

Compare with

we get P = -3 and Q = sin2x

This is linear differential equation where P and Q are functions of x

For the solution of linear differential equation, we first need to find the integrating factor

⇒ IF = e^∫Pdx

⇒ IF = e^∫(-3)dx

⇒ IF = e^-3x

The solution of linear differential equation is given by y(IF) = ∫Q(IF)dx + c

Substituting values for Q and IF

⇒ ye^-3x = ∫e^-3xsin2x dx …. (1)

Let I = ∫e^-3xsin2x dx

If u(x) and v(x) are two functions then by integration by parts,

Here v = sin 2x and u = e^-3x

Applying the above formula, we get,

Again, applying the above stated rule in we get

So,

Put this value in (1) to get,

ye^-3x = ∫e^-3xsin2x dx

Slope of tangent is given as

Slope of tangent of a curve y = f(x) is given by

Put y = vx

Differentiate vx using product rule,

Integrate

Put 1 – v² = t hence differentiating with respect to v we get which means 2vdv = -dt

Resubstitute t

Resubstitute v

Now it is given that the curve is passing through (2, 1)

Hence (2, 1) will satisfy the curve equation (a)

Putting values x = 2 and y = 1 in (a)

Using log a + log b = loga b

Put c in equation (a)

Using log a – log b = log

Using log a + log b = log ab

⇒ 3x = 2(x² – y²)

⇒ 3x = 2x² – 2y²

⇒ 2y² = 2x² – 3x

Hence equation of curve is

Slope of tangent is given as

Slope of tangent of a curve y = f(x) is given by

Integrate

Using partial fraction for

Equating numerator

⇒ A(x + 1) + Bx = 1

Put x = 0

⇒ A = 1

Put x = -1

⇒ B = -1

Hence

Hence equation (a) becomes

⇒ log(y – 1) = logx – log(x + 1) + c …(b)

Now it is given that the curve is passing through (1, 0)

Hence (1, 0) will satisfy the curve equation (b)

Putting values x = 1 and y = 0 in (b)

If we put y = 0 in (b) we get log (-1) which is not defined hence we must simplify further equation (b)

⇒ log (y – 1) – log x = – log (x + 1) + c

Using log a – log b = log

Using log a + log b = log ab

Take the constant c as log c so that we don’t have any undefined terms in our equation (Why only log c and not any other term because taking log c completely eliminates the log terms so we don’t have to worry about the undefined terms appearing in our equation)

Eliminating log

Now substitute x = 1 and y = 0

⇒ c = -2

Put back c = -2 in (c)

Hence the equation of curve is (y – 1)(x + 1) = -2x

Abscissa means the x-coordinate and ordinate means the y-coordinate

Slope of tangent is given as square of the difference of the abscissa and ordinate

Difference of abscissa and ordinate is (x – y) and its square will be (x – y)²

Hence slope of tangent is (x – y)²

Slope of tangent of a curve y = f(x) is given by

Substitute x – y = z hence y = x – z

Differentiate x – z with respect to x

Using partial fraction for

Equating numerator

⇒ A(1 – z) + B(1 + z) = 1

Put z = 1

⇒ B = 1/2

Put z = -1

⇒ A = 1/2

Hence

Put in (a)

Integrate

⇒ x = 1/2(log(1 + z) + (–log(1 – z)) + c

⇒ 2x = log(1 + z) – log(1 – z) + c

Using log a – log b = log

Resubstitute z = x – y

Now it is given that the curve is passing through origin that is (0, 0)

Hence (0, 0) will satisfy the curve equation (b)

Putting values x = 0 and y = 0 in (b)

⇒ c = 0

Put c = 0 back in equation (b)

⇒ e^2x(1 – x + y) = (1 + x – y)

Hence equation of curve is e^2x(1 – x + y) = (1 + x – y)

A(0, a) and B(b, 0) are points on the Y-axis and X-axis respectively and P(x, y) is the midpoint of AB

Now the x-coordinate of the midpoint will be the addition of x coordinates of the points A and B divided by 2

⇒ b = 2x

Similarly, the y-coordinate

⇒ a = 2y

Hence coordinates of A and B are (0, 2y) and (2x, 0) respectively

Now AB is given as tangent to curve having P as point of contact

Slope of line given two points (x₁, y₁) and (x₂, y₂) on it is given by

Here (x₁, y₁) and (x₂, y₂) are A (0, 2y) and B(2x, 0) respectively

⇒ slope of tangent AB =

Hence slope of tangent is

Slope of tangent of a curve y = f(x) is given by

Integrate

⇒ log y = - log x + c

⇒ log y + log x = c

Using log a + log b = log ab

⇒ log xy = c …(a)

Now it is given that the curve is passing through (1, a)

Hence (1, 1) will satisfy the curve equation (a)

Putting values x = 1 and y = 1 in (a)

⇒ log1 = c

⇒ c = 0

Put c = 0 back in (a)

⇒ log xy = 0

⇒ xy = e⁰

⇒ xy = 1

Hence the equation of the curve is xy = 1

Using log a – log b = log

Put y = vx

Differentiate vx with respect to x using product rule

Integrate

Substitute log v = t

Differentiate with respect to v which means

⇒ log t = log x + log c

Resubstitute value of t

⇒ log(log v) = log x + log c

Resubstitute v

Hence solution of given differential equation is

In general terms for a polynomial the degree is the highest power

Degree of differential equation is defined as the highest integer power of highest order derivative in the equation

Here the differential equation is

Now for degree to exist the given differential equation must be a polynomial in some differentials

Here differentials mean

The given differential equation is not polynomial because of the term and hence degree of such a differential equation is not defined.

In general terms for a polynomial the degree is the highest power

The differential equation is

Square both the sides

Now for degree to exist the given differential equation must be a polynomial in some differentials

Here differentials mean

The given differential equation is polynomial in differentials

Degree of differential equation is defined as the highest integer power of highest order derivative in the equation

Here the highest derivative is and there is only one term of highest order derivative in the equation which is whose power is 2 hence degree is 2

The differential equation is

Order is defined as the number which represents the highest derivative in a differential equation

Here is the highest derivative in given equation which is second order hence order of given differential equation is 2

Now let us find the degree

Let us first bring integer powers on the differentials

Take power 4 on both sides

Now for degree to exist the differential equation (a) must be a polynomial in some differentials

Here differentials means

The given differential equation is polynomial in differentials

Degree of differential equation is defined as the highest integer power of highest order derivative in the equation

Observe that in the term of differential equation (a) the maximum power of will be 4

Highest order is and highest power to it is 4

Hence degree of given differential equation is 4

Hence order 2 and degree 4

If y = f(x) is solution of a differential equation then differentiating y = f(x) will give the same differential equation

Let us find the differential equation by differentiating y with respect to x twice

Why twice because all the options have order as 2 and also because there are two constants A and B

y = e^-x (A cos x + B sin x)

Differentiating using product rule

But e^-x (A cos x + B sin x) = y

Differentiating again with respect to x

But e^-x (A cos x + B sin x) = y

Also which means

Let us find the differential equation by differentiating y with respect to x twice

Why twice because we have to eliminate two constants A and B

y = A cos αx + B sin αx

Differentiating

Differentiating again

But y = A cos αx + B sin αx

Let us solve the differential equation

⇒ xdy – ydx = 0

⇒ xdy = ydx

⇒ log y = log x + c

⇒ log y – log x = c

Using log a – log b = log

⇒ y = e^cx

e^c is a constant because e is a constant and c is the integration constant let it be denoted as k hence e^c = k

⇒ y = kx

The equation y = kx is equation of a straight line and (0, 0) satisfies the equation hence

The differential equation is

Compare with we get P = tanx and Q = sec x

The IF integrating factor is given by e^∫Pdx

⇒ e^∫Pdx = e^∫tanxdx

Substitute cosx = t hence hence sinxdx = -dt

⇒ e^∫Pdx = e^-logt

Resubstitute value of t

⇒ e^∫Pdx = e^-log(cosx)

⇒ e^∫Pdx = e^log(secx)

⇒ e^∫Pdx = sec x

Hence IF is sec x

The given differential equation is

⇒ tanysec²xdx + tanxsec²ydy = 0

Divide throughout by tanxtany

Integrate

Put tanx = t hence that is sec²xdx = dt

Put tany = z hence that is sec²ydy = dt

⇒ log t + log z + c = 0

Resubstitute t and z

⇒ log(tan x) + log(tan y) + c = 0

Using log a + log b = log ab

⇒ log(tan x tan y) = -c

⇒ tan x tan y = e^-c

e is a constant -c is a constant hence e^-c is a constant, let it be denoted as k hence k = e^-c

⇒ tan x tan y = k

y = Ax + A³

Let us find the differential equation representing it so we have to eliminate the constant A

Differentiate with respect to x

Put back value of A in y

Now for degree to exist the differential equation must be a polynomial in some differentials

Here differentials mean

The given differential equation is polynomial in differentials

Degree of differential equation is defined as the highest integer power of highest order derivative in the equation

Highest derivative is and highest power to it is 3

Hence degree is 3

Given differential equation

Divide throughout by x

Compare with we get and Q = x³ – 3

The IF integrating factor is given by e^∫Pdx

⇒ e^∫Pdx = e^-logx

Hence IF integrating factor is

Integrate

⇒ log(1 + y) = x + c …(a)

Now it is given that y(0) = 1 which means when x = 0 y = 1

Hence substitute x = 0 and y = 1 in (a)

⇒ log(1 + y) = x + c

⇒ log(1 + 1) = 0 + c

⇒ c = log2

Put c = log2 back in (a)

⇒ log(1 + y) = x + log2

⇒ log(1 + y) – log2 = x

Using log a – log b = log

⇒ 1 + y = 2e^x

⇒ y = 2e^x – 1

Hence solution of differential equation is y = 2e^x – 1

Integrate

⇒ log(y + 1) = log(x – 1) -log c

⇒ log(y + 1) + log c = log(x – 1)

Using log a + log b = log ab

Now it is given that y(1) = 2 which means when x = 1, y = 2

Substitute x = 1 and y = 2 in (a)

⇒ c =0

⇒ x – 1 = 0

So only one solution exists.

Order is defined as the number which represents the highest derivative in a differential equation

Second order means the order should be 2 which means the highest derivative in the equation should be or y’’

Let us examine each option given

A. (y’)² + x = y²

The highest order derivate is y’ which is first order.

B. y’y’’ + y = sinx

The highest order derivate is y’’ which is second order.

C. y’’’+(y’’)²+y=0

The highest order derivate is y’’’ which is third order.

D. y’ = y²

The highest order derivate is y’ which is first order.

Divide throughout by (1 – x²)

Compare with

we get and

The IF integrating factor is given by e^∫Pdx

Substitute 1 – x² = t hence

which means

Hence

⇒ e^∫Pdx = e^log√t

⇒ e^∫Pdx = √t

Resubstitute t

Hence the IF integrating factor is

If y = f(x) is solution of a differential equation, then differentiating y = f(x) will give the same differential equation

Let us find the differential equation by differentiating y with respect to x

⇒ tan^-1x + tan^-1y = c

Differentiating with respect to x

⇒ (1 + y²)dx + (1 + x²)dy = 0

⇒ y dy = (c – x)dx

Integrate

⇒ y dy = (c – x)dx

⇒ ∫y dy = ∫(c – x)dx

⇒ ∫y dy = ∫c dx – ∫x dx

k is the integration constant

⇒ x² + y² = 2cx + k

⇒ x² + y² – 2cx – k = 0

This is equation of circle because there is no ‘xy’ term and the coefficient of x² and y² is same.

This equation represents family of circles because for different values of c and k we will get different circles

e^xcos y dx – e^xsin y dy = 0

⇒ e^xcos y dx = e^xsin y dy

Integrate

Substitute cosy = t hence which means siny dy = -dt

⇒ x = -log t + c

Resubstitute t

⇒ x = -log(cosy) + c

⇒ x + c = log(cosy)^-1

⇒ x + c = log(secy)

⇒ e^x+c = sec y

⇒ e^x × e^c = sec y

⇒ e^xcos y = e^-c

As e is a constant c is the integration constant hence e^-c is a constant and hence let it be denoted by k such that k = e^-c

⇒ e^xcos y = k