# 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^{2x}dx + cy(e

^{2x}) = (1/5)e^{5x }+ cy = (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^{2x}dx + cy(e

^{2x}) = (5/4)∫e^{-x}dx + cy = -(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^{2x}dx + cy(e

^{2x}) = 6∫e^{3x}dx + cy(e

^{2x}) = 2e^{3x }+ cy = 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^{x}dx + cy(e

^{x}) = ∫e^{-x}dx + cy(e

^{x}) = -e^{-x }+ cy = -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^{2x}dx + cy(e

^{2x}) = 4x∫e^{2x}dx – 4∫{(dx/dx)∫e^{2x}dx}dx + cy(e

^{2x}) = 2xe^{2x }– 2∫e^{2x}dx + cy(e

^{2x}) = 2xe^{2x }– e^{2x }+ cy = (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

^{x}dx + cxy = x∫e

^{x}dx – {(dx/dx)∫e^{x}dx}dx + cxy = xe

^{x }– ∫e^{x}dx + cxy = xe

^{x }– e^{x }+ cThis is the required solution.

### Question 8. (dy/dx) + [4x/(x^{2 }+ 1)]y = -1/(x^{2 }+ 1)^{2}

**Solution**:

We have,

(dy/dx) + [4x/(x

^{2 }+ 1)]y = -1/(x^{2 }+ 1)^{2}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = [4x/(x

^{2 }+ 1)], Q = -1/(x^{2 }+ 1)^{2}So,

I.F = e

^{∫Pdx}= (x

^{2}+1)^{2}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(x

^{2 }+ 1)^{2 }= ∫-[1/(x^{2 }+ 1)^{2}](x^{2 }+ 1)^{2}dx + cy(x

^{2 }+ 1)^{2 }= -∫dx + cy(x

^{2 }+ 1)^{2 }= -x + cy = -x/(x

^{2 }+ 1)^{2 }+ c/(x^{2 }+ 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 = (x

^{2}/2)logx – ∫(1/x)(x^{2}/2) + cxy = (x

^{2}/2)logx – (1/2)∫xdx + cxy = (x

^{2}/2)logx – (x^{2}/4) + cy = (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/x) = ∫[(1/x) – (1/x

^{2})]e^{x }+ cSince, ∫[f(x) + f'(x)]e

^{x}dx = f(x)e^{x }+ c(y/x) = (e

^{x}/x) + cy = e

^{x }+ xcThis is the required solution.

### Question 11. (dy/dx) + y/x = x^{3}

**Solution:**

We have,

(dy/dx) + y/x = x

^{3}………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/x, Q = x

^{3}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 = ∫x

^{3}.xdx + cyx = ∫x

^{4}dx + cyx = (x

^{5}/5) + cy = (x

^{4}/5) + c/xThis 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^{x}dx + cLet, I = ∫sinx.e

^{x}dxI = e

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

^{x}cos + ∫e^{x}cosxdxI = -e

^{x}cosx + e^{x}∫cosxdx – ∫{(d/dx)e^{x}∫cosxdx}dxI = -e

^{x}cosx + e^{x}sinx – ∫e^{x}sinxdx2I = e

^{x}(sinx – cosx)I = (e

^{x}/2)(sinx – cosx)y(e

^{x}) = (e^{x}/2)(sinx – cosx) + cy = (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^{x}dx + cLet, I = ∫cosx.e

^{x}dxI = e

^{x}∫cosxdx – ∫{(d/dx)e^{x}∫cosxdx}dxI = e

^{x}sinx – ∫e^{x}sinxdxI = e

^{x}sinx – e^{x}∫sinxdx + ∫{(d/dx)e^{x}∫sinxdx}dxI = e

^{x}sinx + e^{x}cosx – ∫e^{x}cosxdx2I = e

^{x}(cosx + sinx)I = (e

^{x}/2)(cosx + sinx)y(e

^{x}) = (e^{x}/2)(cosx + sinx) + cy = (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^{2x}dx + cLet, I = ∫sinx.e

^{2x}dxI = e

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

^{2x}cosx + 2∫e^{2x}cosdxI = -e

^{2x}cosx + 2e^{2x}∫cosxdx – 2{(d/dx)e^{2x}∫cosxdx}dxI = -e

^{2x}cosx + 2e^{2x}sinx – 4∫e^{2x}sinxdx5I = e

^{2x}(2sinx – cosx)I = (e

^{2x}/5)(2sinx – cosx)y(e

^{2x}) = (e^{2x}/5)(2sinx – cosx) + cy = (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 + x^{2})(dy/dx) + y = tan^{-1}x

**Solution:**

We have,

(1 + x

^{2})(dy/dx) + y = tan^{-1}x(dy/dx) + [y/(1 + x

^{2})] = tan^{-1}x/(1 + x^{2}) ………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(1 + x

^{2}), Q = tan^{-1}x/(1 + x^{2})So,

I.F = e

^{∫Pdx}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

Let, tan

^{-1}x = zOn differentiating both sides we get,

dx/(1 + x

^{2}) = dzy(e

^{z}) = ∫ze^{z}dz + cy(e

^{z}) = z∫e^{z}dz – {(dz/dz)∫e^{z}dz}dzy(e

^{z}) = ze^{z }– ∫ e^{z}dz + cy(e

^{z}) = e^{z}(z – 1) + cy = (z – 1) + ce

^{-z}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 = x^{2}cotx + 2x

**Solution:**

We have,

(dy/dx) + ycotx = x

^{2}cotx + 2x ………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = cotx, Q = x

^{2}cotx + 2xSo,

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) = ∫(x

^{2}cotx + 2x)sinxdx + cy(sinx) = ∫(x

^{2}cosx + 2xsinx)dx + cy(sinx) = x

^{2}∫cosxdx – {(d/dx)x^{2}∫cosxdx}dx + ∫2xsinxdx + cy(sinx) = x

^{2}sinx – ∫2xsinxdx + ∫2xsinxdx + cy(sinx) = x

^{2}sinxdx + cThis is the required solution.

### Question 19. (dy/dx) + ytanx = x^{2}cos^{2}x

**Solution:**

We have,

(dy/dx) + ytanx = x

^{2}cos^{2}x ………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = tanx, Q = x

^{2}cos^{2}xSo,

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.(x

^{2}.cos^{2}x)dx + cy(secx) = ∫x

^{2}.cosxdx + cy(secx) = x

^{2}∫cosxdx – ∫{(d/dx)x^{2}∫cosxdx}dx + cy(secx) = x

^{2}sinx – 2∫xsinxdx + cy(secx) = x

^{2}sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + cy(secx) = x

^{2}sinx + 2xcosx – 2∫cosxdx + cy(secx) = x

^{2}sinx + 2xcosx – 2sinx + cThis is the required solution.

**Question 20.** (1 + x^{2})(dy/dx) + y =

**Solution:**

We have,

(1 + x

^{2})(dy/dx) + y =(dy/dx) + [1/(x

^{2 }+ 1)]y = /(x^{2 }+ 1) ………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = 1/(x

^{2 }+ 1), Q = /(x^{2}+ 1)So,

I.F = e

^{∫Pdx}=

The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y() – ∫[/(x

^{2 }+ 1)]dx + cLet, tan

^{-1}x = zOn differentiating both sides we get

dx/(1 + x

^{2}) = dzye

^{z }= ∫e^{2z}dz + cye

^{z }= (e^{2z}/2) + cy = (e

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

^{ }+ c.This is the required solution.

### Question 21. xdy = (2y + 2x^{4 }+ x^{2})dx

**Solution:**

We have,

xdy = (2y + 2x

^{4 }+ x^{2})dx(dy/dx) = 2(y/x) + 2x

^{3 }+ x(dy/dx) – (y/x) = 2x

^{3 }+ x ………..(i)The given equation is a linear differential equation of the form

(dy/dx) + Py = Q

Where, P = -(2/x), Q = 2x

^{3 }+ xSo,

I.F = e

^{∫Pdx}= e

^{-2∫(dx/x)}= e

^{-2log(x)}= e

^{2log|1/x|}= 1/x

^{2}The solution of a differential equation is,

y(I.F) = ∫Q(I.F)dx + c

y(1/x

^{2}) = ∫(1/x^{2}).(2x^{3 }+ x)dx + c(y/x

^{2}) = ∫[2x + (1/x)]dx + c(y/x

^{2}) = x^{2 }+ log|x| + cy = x

^{4 }+ x^{2}log|x| + cx^{2}This is the required solution.

### Question 22. (1 + y^{2}) + (x – )(dy/dx) = 0

**Solution:**

We have,

(1 + y

^{2}) + (x – )(dy/dx) = 0(dx/dy) + x/(1 + y

^{2}) = /(1 + y^{2}) ………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/(1 + y

^{2}), Q = /(1 + y^{2})So,

I.F = e

^{∫Pdy}=

The solution of a differential equation is,

x(I.F) = ∫Q(I.F)dy + c

Let, tan

^{-1}y = zOn differentiating both sides we have,

dy/(1 + y

^{2}) = dzxe

^{z }= ∫e^{2z}dz + cxez = (e2z/2) + c

x = e

^{z}/2 + ce^{-z}This is the required solution.

### Question 23. y^{2}(dx/dy) + x – 1/y = 0

**Solution:**

We have,

y

^{2}(dx/dy) + x – 1/y = 0(dx/dy) + (x/y

^{2}) = 1/y^{3}………..(i)The given equation is a linear differential equation of the form

(dx/dy) + Px = Q

Where, P = 1/y

^{2}, Q = 1/y^{3}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/y^{3}).(e^{-1/y})dy + cLet,-(1/y) = z

Differentiating both sides we have,

(dy/y

^{2}) = dzxe

^{z }= -∫ze^{z}dz + cxe

^{z }= -z∫e^{z}dz + ∫{(dz/dz)∫e^{z}dz}dz + cxe

^{z }= -ze^{z }+ ∫e^{z}dz + cxe

^{z }= -ze^{z }+ e^{z }+ cx = (1 – z) + ce – z

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

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

^{1/y}This is the required solution.

Attention reader! Don’t stop learning now. Join the **First-Step-to-DSA Course for Class 9 to 12 students ****, **specifically designed to introduce data structures and algorithms to the class 9 to 12 students