Class 12 RD Sharma Solutions - Chapter 22 Differential Equations - Exercise 22.10 | Set 1

Solve the following differential equations

Question 1. dy/dx + 2y = e^3x

Solution:

We have,

dy/dx + 2y = e^3x  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = e^3x

So, I.F = e^∫Pdx

= e^∫2dx

= e^2x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = ∫e^3x.e^2xdx + c

y(e^2x) = (1/5)e^5x + c

y = (e^3x/5) + ce^-2x

Hence, this is the required solution.

Question 2. 4(dy/dx) + 8y = 5e^-3x

Solution:

We have,

4(dy/dx) + 8y = 5e^-3x 

(dy/dx) + 2y = (5/4)e^-3x  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = (5/4)e^-3x

So, I.F = e^∫Pdx

= e^∫2dx

= e^2x

The solution of differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = (5/4)∫e^-3x.e^2xdx + c

y(e^2x) = (5/4)∫e^-xdx + c

y = -(5/4)e^-3x + ce^-2x

This is the required solution.

Question 3. (dy/dx) + 2y = 6e^x

Solution:

We have,

(dy/dx) + 2y = 6e^x ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = 6e^x

So, I.F = e^∫Pdx

= e^∫2dx

= e^2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = ∫6e^x.e^2xdx + c

y(e^2x) = 6∫e^3xdx + c

y(e^2x) = 2e^3x + c

y = 2e^x + ce^-2x

This is the required solution.

Question 4. (dy/dx) + y = e^-2x

Solution:

We have,

(dy/dx) + y = e^-2x ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = e^-2x

So, I.F = e^∫Pdx

= e^∫dx

= e^x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^x) = ∫e^-2x.e^xdx + c

y(e^x) = ∫e^-xdx + c

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

y = -e^-2x + ce^-x

This is the required solution.

Question 5. x(dy/dx) = x + y

Solution:

We have,

x(dy/dx) = x + y

(dy/dx) = 1 + (y/x)

(dy/dx) - (y/x) = 1  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (-1/x), Q = 1

So, I.F = e^∫Pdx

= e^-∫(dx/x)

= e^-log(x)

= e^log(1/x)

= (1/x)

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x) = ∫(1/x)dx + c

(y/x) = log|x| + c

y = xlog|x| + cx

This is the required solution.

Question 6. (dy/dx) + 2y = 4x

Solution:

We have,

(dy/dx) + 2y = 4x   ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = 4x

So,

I.F = e^∫Pdx

= e^∫2dx

= e^2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = ∫4x.e^2xdx + c

y(e^2x) = 4x∫e^2xdx - 4∫{(dx/dx)∫e^2xdx}dx + c

y(e^2x) = 2xe^2x - 2∫e^2xdx + c

y(e^2x) = 2xe^2x - e^2x + c

y = (2x - 1) + ce^-2x

This is the required solution.

Question 7. x(dy/dx) + y = xe^x

Solution:

We have,

x(dy/dx) + y = xe^x

(dy/dx) + (y/x) = e^x  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (1/x), Q = e^x

So,

I.F = e^∫Pdx

= e^∫(dx/x)

= e^log(x)

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫x.e^xdx + c

xy = x∫e^xdx - {(dx/dx)∫e^xdx}dx + c

xy = xe^x - ∫e^xdx + c

xy = xe^x - e^x + c

y=\frac{x-1}{x}e^x+\frac{c}{x}

This is the required solution.

Question 8. (dy/dx) + [4x/(x² + 1)]y = -1/(x² + 1)²

Solution:

We have,

(dy/dx) + [4x/(x² + 1)]y = -1/(x² + 1)²    ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = [4x/(x² + 1)], Q = -1/(x² + 1)²

So,

I.F = e^∫Pdx

=e^{∫\frac{4x}{x^2+1}dx}

=e^{2log|x^2+1|}

= (x²+1)²

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x² + 1)² = ∫-[1/(x² + 1)²](x² + 1)²dx + c

y(x² + 1)² = -∫dx + c

y(x² + 1)² = -x + c

y = -x/(x² + 1)² + c/(x² + 1)²

This is the required solution.

Question 9. x(dy/dx) + y = xlogx

Solution:

We have,

x(dy/dx) + y = xlogx

(dy/dx) + (y/x) = logx  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = (1/x), Q = logx

So,

I.F = e^∫Pdx

= e^∫(dx/x)

= e^log(x)

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x) = ∫x.logxdx + c

xy = logx∫xdx - {(d/dx)logx∫xdx}dx + c

xy = (x²/2)logx - ∫(1/x)(x²/2) + c

xy = (x²/2)logx - (1/2)∫xdx + c

xy = (x²/2)logx - (x²/4) + c

y = (x/2)logx - (x/4) + (c/x)

This is the required solution.

Question 10. x(dy/dx) - y = (x - 1)e^x

Solution:

We have,

x(dy/dx) - y = (x - 1)e^x

(dy/dx) - (y/x) = [(x - 1)/x]e^x  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(1/x), Q = [(x - 1)/x]e^x

So,

I.F = e^∫Pdx

= e^-∫(dx/x)

= e^-log(x)

= e^log(1/x)

= 1/x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(\frac{1}{x})=∫(\frac{x-1}{x})e^x.(\frac{dx}{x})+c

(y/x) = ∫[(1/x) - (1/x²)]e^x + c

Since, ∫[f(x) + f'(x)]e^xdx = f(x)e^x + c

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

y = e^x + xc

This is the required solution.

Question 11. (dy/dx) + y/x = x³

Solution:

We have,

(dy/dx) + y/x = x³     ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = x³

So, I.F = e^∫Pdx

= e^∫dx/x

= e^logx

= x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

yx = ∫x³.xdx + c

yx = ∫x⁴dx + c

yx = (x⁵/5) + c

y = (x⁴/5) + c/x

This is the required solution.

Question 12. (dy/dx) + y = sinx

Solution:

We have,

(dy/dx) + y = sinx ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = sinx

So, I.F = e^∫Pdx

= e^∫dx

= e^x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^x) = ∫sinx.e^xdx + c

Let, I = ∫sinx.e^xdx

I = e^x∫sinxdx - ∫{(d/dx)e^x∫sinxdx}dx

I = -e^xcos + ∫e^xcosxdx

I = -e^xcosx + e^x∫cosxdx - ∫{(d/dx)e^x∫cosxdx}dx

I = -e^xcosx + e^xsinx - ∫e^xsinxdx

2I = e^x(sinx - cosx)

I = (e^x/2)(sinx - cosx)

y(e^x) = (e^x/2)(sinx - cosx) + c

y = (1/2)(sinx - cosx) + ce^-x

This is the required solution.

Question 13. (dy/dx) + y = cosx

Solution:

We have,

(dy/dx) + y = cosx ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1, Q = cosx

So, I.F = e^∫Pdx

= e^∫dx

= e^x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^x) = ∫cosx.e^xdx + c

Let, I = ∫cosx.e^xdx

I = e^x∫cosxdx - ∫{(d/dx)e^x∫cosxdx}dx

I = e^xsinx - ∫e^xsinxdx

I = e^xsinx - e^x∫sinxdx + ∫{(d/dx)e^x∫sinxdx}dx

I = e^xsinx + e^xcosx - ∫e^xcosxdx

2I = e^x(cosx + sinx)

I = (e^x/2)(cosx + sinx)

y(e^x) = (e^x/2)(cosx + sinx) + c

y = (1/2)(cosx + sinx) + ce^-x

This is the required solution.

Question 14. (dy/dx) + 2y = sinx

Solution:

We have,

(dy/dx) + 2y = sinx ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 2, Q = sinx

So, I.F = e^∫Pdx

= e^∫2dx

= e^2x

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^2x) = ∫sinx.e^2xdx + c

Let, I = ∫sinx.e^2xdx

I = e^2x∫sinxdx - {(d/dx)e^2x∫sinxdx}dx

I = -e^2xcosx + 2∫e^2xcosdx

I = -e^2xcosx + 2e^2x∫cosxdx - 2{(d/dx)e^2x∫cosxdx}dx

I = -e^2xcosx + 2e^2xsinx - 4∫e^2xsinxdx

5I = e^2x(2sinx - cosx)

I = (e^2x/5)(2sinx - cosx)

y(e^2x) = (e^2x/5)(2sinx - cosx) + c

y = (1/5)(2sinx - cosx) + ce^-2x

This is the required solution.

Question 15. (dy/dx) - ytanx = -2sinx

Solution:

We have,

(dy/dx) - ytanx = -2sinx ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -tanx, Q = sinx

So, 

I.F = e^∫Pdx

= e^∫-tanxdx

= e^-log|secx|

= 1/secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/secx) = -2∫sinx.(1/secx)dx + c

ycosx = -∫2sinx.cosxdx + c

ycosx = -∫sin2xdx + c

ycosx = (cos2x/2) + c

y = (cos2x/cosx) + (c/cosx)

This is the required solution.

Question 16. (1 + x²)(dy/dx) + y = tan^-1x

Solution:

We have,

(1 + x²)(dy/dx) + y = tan^-1x

(dy/dx) + [y/(1 + x²)] = tan^-1x/(1 + x²)  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(1 + x²), Q = tan^-1x/(1 + x²)

So,

I.F = e^∫Pdx

=e^{∫\frac{dx}{1+x^2}}

=e^{tan^{-1}x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{tan^{-1}x})=∫\frac{tan^{-1}x}{1+x^2}e^{tan^{-1}x}dx+c

Let, tan^-1x = z

On differentiating both sides we get,

dx/(1 + x²) = dz

y(e^z) = ∫ze^zdz + c

y(e^z) = z∫e^zdz - {(dz/dz)∫e^zdz}dz

y(e^z) = ze^z - ∫ e^zdz + c

y(e^z) = e^z(z - 1) + c

y = (z - 1) + ce^-z

y=(tan^{-1}x-1)+ce^{tan^{-1}x}

This is the required solution.

Question 17. (dy/dx) + ytanx = cosx

Solution:

We have,

(dy/dx) + ytanx = cosx  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = cosx

So, I.F = e^∫Pdx

= e^∫tanxdx

= e^log|secx|

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y.secx = ∫cosx.secxdx + c

y.secx = ∫dx + c

y.secx = x + c

y = xcosx + c.cosx

This is the required solution.

Question 18. (dy/dx) + ycotx = x²cotx + 2x

Solution:

We have,

(dy/dx) + ycotx = x²cotx + 2x ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = x²cotx + 2x

So,

I.F = e^∫Pdx

= e^∫cotxdx

= e^log|sinx|

= sinx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(sinx) = ∫(x²cotx + 2x)sinxdx + c

y(sinx) = ∫(x²cosx + 2xsinx)dx + c

y(sinx) = x²∫cosxdx - {(d/dx)x²∫cosxdx}dx + ∫2xsinxdx + c

y(sinx) = x²sinx - ∫2xsinxdx + ∫2xsinxdx + c

y(sinx) = x²sinxdx + c

This is the required solution.

Question 19. (dy/dx) + ytanx = x²cos²x

Solution:

We have,

(dy/dx) + ytanx = x²cos²x  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x²cos²x

So,

I.F = e^∫Pdx

= e^∫tanxdx

= e^log|secx|

= secx

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(secx) = ∫secx.(x².cos²x)dx + c

y(secx) = ∫x².cosxdx + c

y(secx) = x²∫cosxdx - ∫{(d/dx)x²∫cosxdx}dx + c

y(secx) = x²sinx - 2∫xsinxdx + c

y(secx) = x²sinx - 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c

y(secx) = x²sinx + 2xcosx - 2∫cosxdx + c

y(secx) = x²sinx + 2xcosx - 2sinx + c

This is the required solution.

Question 20. (1 + x²)(dy/dx) + y = e^{tan^{-1}x}

Solution:

We have,

(1 + x²)(dy/dx) + y = e^{tan^{-1}x}

(dy/dx) + [1/(x² + 1)]y = e^{tan^{-1}x}/(x² + 1)  ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(x² + 1), Q = e^{tan^{-1}x}/(x² + 1)

So,

I.F = e^∫Pdx

=e^{∫\frac{1}{x^2+1}dx}

= e^{tan^{-1}x}

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(e^{tan^{-1}x}) - ∫[e^{tan^{-1}x}/(x² + 1)]e^{tan^{-1}x}dx + c

Let, tan^-1x = z

On differentiating both sides we get

dx/(1 + x²) = dz

ye^z = ∫e^2zdz + c

ye^z = (e^2z/2) + c

y = (e^z/z) + c.e^-z

y = (1/2)e^{tan^{-1}x} + c.e^{tan^{-1}x}

This is the required solution.

Question 21. xdy = (2y + 2x⁴ + x²)dx

Solution:

We have,

xdy = (2y + 2x⁴ + x²)dx

(dy/dx) = 2(y/x) + 2x³ + x

(dy/dx) - (y/x) = 2x³ + x   ...........(i)

The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(2/x), Q = 2x³ + x

So,

I.F = e^∫Pdx

= e^-2∫(dx/x)

= e^-2log(x)

= e^2log|1/x|

= 1/x²

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x²) = ∫(1/x²).(2x³ + x)dx + c

(y/x²) = ∫[2x + (1/x)]dx + c

(y/x²) = x² + log|x| + c

y = x⁴ + x²log|x| + cx²

This is the required solution.

Question 22. (1 + y²) + (x - e^{tan^{-1}y})(dy/dx) = 0

Solution:

We have,

(1 + y²) + (x - e^{tan^{-1}y})(dy/dx) = 0

\frac{dy}{dx}=-\frac{1+y^2}{x-e^{tan^{-1}}x}

\frac{dx}{dy}=-\frac{x-e^{tan^{-1}}x}{1+y^2}

(dx/dy) + x/(1 + y²) = e^{tan^{-1}y}/(1 + y²)  ...........(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/(1 + y²), Q = e^{tan^{-1}y}/(1 + y²)

So,

I.F = e^∫Pdy

e^{∫\frac{dy}{1+y^2}}

= e^{tan^{-1}y}

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

x(e^{tan^{-1}y})=∫(e^{tan^{-1}y})(\frac{e^{tan^{-1}y}}{1+y^2})dy+c

Let, tan^-1y = z

On differentiating both sides we have,

dy/(1 + y²) = dz

xe^z = ∫e^2zdz + c

xez = (e2z/2) + c

x = e^z/2 + ce^-z

x=\frac{e^{tan^{-1}y}}{2}+c.e^{tan^{-1}y}

This is the required solution.

Question 23. y²(dx/dy) + x - 1/y = 0

Solution:

We have,

y²(dx/dy) + x - 1/y = 0

(dx/dy) + (x/y²) = 1/y³    ...........(i)

The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y², Q = 1/y³

So,

I.F = e^∫Pdy

= e^∫dy/y2

= e^-(1/y)

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

e^-(1/y)x = ∫(1/y³).(e^-1/y)dy + c

Let,-(1/y) = z

Differentiating both sides we have,

(dy/y²) = dz 

xe^z = -∫ze^zdz + c

xe^z = -z∫e^zdz + ∫{(dz/dz)∫e^zdz}dz + c

xe^z = -ze^z + ∫e^zdz + c

xe^z = -ze^z + e^z + c

x = (1 - z) + ce - z

x = [1 + (1/y)] + ce^1/y

x = (y + 1)/y + ce^1/y

This is the required solution.

Summary

Exercise 22.10 in Chapter 22 typically deals with second-order linear differential equations. These are equations of the form:

a(x)y'' + b(x)y' + c(x)y = f(x)

where a(x), b(x), c(x), and f(x) are functions of x, and y is a function of x. The prime notation denotes derivatives with respect to x.

Key points:

1. If f(x) = 0, the equation is called homogeneous; otherwise, it's non-homogeneous.

2. The general solution of a second-order linear differential equation is the sum of the complementary function (CF) and the particular integral (PI).

3. The CF is the general solution of the corresponding homogeneous equation.

4. The PI is any solution of the non-homogeneous equation.

5. Various methods are used to solve these equations, including the method of undetermined coefficients and variation of parameters.

Practice Questions

1. Solve the differential equation: y'' - 5y' + 6y = 0

2. Find the general solution of: y'' + 4y' + 4y = e^x

3. Solve: y'' - 2y' - 3y = x^2

4. Find a particular solution of: y'' + y = sin(x)

5. Solve the initial value problem: y'' - y = e^x, y(0) = 1, y'(0) = 0

6. Find the general solution of: y'' + 9y = 18 cos(3x)

7. Solve: y'' + 2y' + y = x e^(-x)

8. Find the complementary function of: y'' - 4y' + 4y = x^2 + 1

9. Solve the differential equation: y'' + 4y = 4 sec^2(2x)

10. Find the particular integral of: y'' - 2y' + y = x^2 + e^x