Here we provide RD Sharma Class 12 Ex 22.5 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.5 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 22 |

Exercise | 22.5 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 22.5 Solutions Chapter 22 Differential Equations**

### Solve the following differential equations:

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

**Solution:**

We have,

(dy/dx) = x

^{2 }+ x – (1/x)dy = [x

^{2 }+ x – (1/x)]dxOn integrating both sides, we get

∫(dy) = ∫[x

^{2 }+ x – (1/x)]dxy = (x

^{3}/3) + (x^{2}/2) – log(x) + c -(Here, ‘c’ is integration constant)

### Question 2. (dy/dx) = x^{5 }+ x^{2 }– (2/x)

**Solution:**

We have,

(dy/dx) = x

^{5 }+ x^{2 }– (2/x)On integrating both sides, we get

∫(dy) = ∫[x

^{5 }+ x^{2 }– (2/x)]dxy = (x

^{6}/6) + (x^{3}/3) – 2log(x) + c -(Here, ‘c’ is integration constant)

### Question 3. (dy/dx) + 2x = e^{3x}

**Solution:**

We have,

(dy/dx) + 2x = e

^{3x}(dy/dx) = -2x + e

^{3x}dy = [-2x + e

^{3x}]dxOn integrating both sides, we get

∫dy = ∫[-2x + e

^{3x}]dxy = (-2x

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

^{2}) + (e^{3x}/3) + c -(Here, ‘c’ is integration constant)

### Question 4. (x^{2 }+ 1)(dy/dx) = 1

**Solution:**

We have,

(x

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

^{2 }+ 1)dxOn integrating both sides, we get

∫dy = ∫(dx/x

^{2 }+ 1)y = tan

^{-1}x + c -(Here, ‘c’ is integration constant)

### Question 5. (dy/dx) = (1 – cosx)/(1 + cosx)

**Solution:**

We have,

(dy/dx) = (1 – cosx)/(1 + cosx)

(dy/dx) = (2sin

^{2}x/2)/(2cos^{2}x/2)dy/dx = tan

^{2}(x/2)dy/dx = [sec

^{2}(x/2) – 1]On integrating both sides, we get

∫dy = ∫[sec

^{2}(x/2) – 1]dxy = [tan(x/2)] – x + c

y = 2tan(x/2) – x + c -(Here, ‘c’ is integration constant)

### Question 6. (x + 2)(dy/dx) = x^{2 }+ 3x + 7

**Solution:**

We have,

(x + 2)(dy/dx) = x

^{2 }+ 3x + 7(dy/dx) = (x

^{2 }+ 3x + 7)/(x + 2)(dy/dx) = (x

^{2 }+ 3x + 2 + 5)/(x + 2)(dy/dx) = [(x + 1)(x + 2) + 5]/(x + 2)

(dy/dx) = (x + 1) + 5/(x + 2)

dy = (x + 1)dx + 5dx/(x + 2)

On integrating both sides, we get

∫dy = ∫(x + 1)dx + 5∫dx/(x + 2)

y = x

^{2}/2 + x + 5log(x + 2) + c -(Here, ‘c’ is integration constant)

### Question 7. (dy/dx) = tan^{-1}x

**Solution:**

We have,

(dy/dx) = tan

^{-1}xdy = tan

^{-1}xdxOn integrating both sides, we get

∫dy = ∫tan

^{-1}xdxy = ∫1 × tan

^{-1}xdxy = tan

^{-1}x∫1dx – ∫[∫1dx]dxy = xtan

^{-1}x – ∫\frac{x}{(1 + x2)}dxy = xtan

^{-1}x –y = xtan

^{-1}x – (1/2)log(1 + x^{2}) + c -(Here, ‘c’ is integration constant)

### Question 8. (dy/dx) = logx

**Solution:**

We have,

(dy/dx) = logx

dy = logxdx

On integrating both sides, we get

∫dy = logx∫1dx – ∫[∫1dx]dx

y = xlogx – ∫[x/x]dx

y = xlogx – ∫dx

y = xlogx – x + c

y = x(logx – 1) + c -(Here, ‘c’ is integration constant)

### Question 9. (1/x)(dy/dx) = tan^{-1}x

**Solution:**

We have,

(1/x)(dy/dx) = tan

^{-1}xdy = xtan

^{-1}x dxOn integrating both sides, we get

∫dy = ∫xtan

^{-1}x dxy = tan

^{-1}x∫xdx – ∫[∫xdx]dxy = (x

^{2}/2)tan^{-1}x –y = (x

^{2}/2)tan^{-1}x – (1/2)∫1- \frac{1}{(1 + x^2)}dxy = (x

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

^{2 }+ 1)tan^{-1}x – (x/2) + c -(Here, ‘c’ is integration constant)

### Question 10. (dy/dx) = cos^{3}xsin^{2}x + x√(2x + 1)

**Solution:**

We have,

(dy/dx) = cos

^{3}xsin^{2}x + x√(2x + 1)(dy/dx) = cosxcos

^{2}xsin^{2}x + x√(2x + 1)y = I

_{1 }+ I_{2}I

_{1 }= cosxcos^{2}xsin^{2}xdxOn integrating both sides, we get

= ∫(1 – sin

^{2}x)sin^{2}xcosxdxLet, sinx = z

On differentiating both sides

cosxdx = dz

= ∫(1 – z

^{2})z^{2}dz= z

^{3}/3 – z^{5}/5 + c_{1}= sin

^{3}x/3 – sin^{5}x/5 + c_{1}I

_{2 }= x√(2x + 1)dxLet, (2x + 1) = u

^{2}On differentiating both sides

2dx = 2udu

dx = udu

= [(u

^{2 }– 1)/2]u × uduI

_{2 }= (1/2)(u^{4}-u^{2})duOn integrating both sides, we get

= (1/2)[u

^{5}/5 – u^{3}/3] + c_{2}= 1/10(2x + 1)

^{5/2}– 1/6(2x + 1)^{3/2 }+ c_{2}y = I

_{1 }+ I_{2}y = (sin

^{3}x/3) – (sin^{5}x/5) + 1/10(2x + 1)^{5/2}– 1/6(2x + 1)^{3/2 }+ c

### Question 11. (sinx + cosx)dy + (cosx – sinx)dx = 0

**Solution:**

We have,

(sinx + cosx)dy + (cosx – sinx)dx = 0

(dy/dx) = -[(cosx – sinx)/(sinx + cosx)]

Let, (sinx + cosx) = z

On differentiating both sides

(sinx + cosx)dx = dz

dy = (dz/z)

On integrating both sides, we get

∫dy = ∫(dz/z)

y = logz + c

y = log(sinx + cosx) + c

### Question12. (dy/dx) – xsin^{2}x = 1/(xlogx)

**Solution:**

We have,

(dy/dx) – xsin

^{2}x = 1/(xlogx)dy = xsin

^{2}xdx + 1/(xlogx)dxOn integrating both sides, we get

y = ∫xsin

^{2}xdx + ∫dx/(xlogx)y = I

_{1 }+ I_{2}I

_{1 }= (1/2)(2sin^{2}x)xdx= (1/2)[(1 – cos2x)xdx]

= (1/2)(xdx – xcos2x)dx

On integrating both sides, we get

= 1/2[∫xdx – ∫xcos2x dx]

= 1/2(x

^{2}/2) – 1/2[x∫cos2x dx – ∫(1∫cos2x dx)]dx= 1/2(x

^{2}/2) – (x/4)sin2x + ∫(1/4sin2x)dx= 1/2(x

^{2}/2) – (x/4)sin2x + ((1/8)cos2x) + c_{1}I

_{2 }= 1/(xlogx)dxLet, logx = z

On differentiating both sides, we get

(dx/x) = dz

= (dz/z)

= logz

= log(logx) + c

_{2}y = I

_{1 }+ I_{2}y = (x

^{2}/4) – (xsin2x/4) + (cos2x/8) + log(logx) + c

### Question 13. (dy/dx) = x^{5}tan^{-1}(x^{3})

**Solution:**

We have,

(dy/dx) = x

^{5}tan^{-1}(x^{3})Let, x

^{3 }= zOn differentiating both sides, we get

3x

^{2}dx = dzx

^{2}dx = dz/3dy = (1/3)[ztan

^{-1}z]dzOn integrating both sides, we get

∫dy = (1/3)∫ztan

^{-1}z dzy = (1/3)[tan

^{ – 1}z ∫zdz – ∫{∫zdz}dz]y = (z

^{2}/6)tan^{-1}z –y = (z

^{2}/6)tan^{-1}z – (1/6)∫1 – \frac{1}{(1 + z^2)}dzy = (z

^{2}/6)tan^{-1}z – (1/6)∫dz – (1/6)∫dz/(1 + z^{2})y = (z

^{2}/6)tan^{-1}z – (z/6) – (1/6)tan^{-1}zy = (1/6)(z

^{2 }+ 1)tan^{-1}z – (z/6) + cy = (1/6)[(x

^{6 }+ 1)tan – 1(x^{3}) – (x^{3})] + c

### Question 14. sin^{4}x(dy/dx) = cosx

**Solution:**

We have,

sin

^{4}x(dy/dx) = cosxdy = (cosx/sin

^{4}x)dxLet, sinx = z

On differentiating both sides, we get

cosx dx = dz

dy = (dz/z

^{4})On integrating both sides, we get

∫(dy) = ∫(1/z

^{4})dzy = (1/ -3t

^{3}) + cy = -(1/3sin

^{3}x) + cy = (-cosec

^{3}x/3) + c -(Here, ‘c’ is integration constant)

### Question 15. cosx(dy/dx) – cos2x = cos3x

**Solution:**

We have,

cosx(dy/dx) – cos2x = cos3x

(dy/dx) = (cos3x + cos2x)/cosx

(dy/dx) = (4cos

^{3}x – 3cosx + 2cos^{2}x – 1)/cosx(dy/dx) = (4cos

^{3}x/cosx) – 3(cosx/cosx) + 2(cos^{2}x/cosx) – secxdy = [4cos

^{2}x – 3 + 2cosx – secx]dxdy = [4{(cos2x + 1)/2} – 3 + 2cosx – secx]dx

On integrating both sides, we get

∫dy = ∫[2cos2x – 1 + 2cosx – secx]dx

y = sin2x – x + 2sinx – log|secx + tanx| + c -(Here, ‘c’ is integration constant)

### Question 16. √(1 – x^{4})(dy/dx) = xdx

**Solution:**

We have,

√(1 – x

^{4})(dy/dx) = xdxLet, x

^{2 }= zOn differentiating both sides, we get

2xdx = dz

xdx = (dz/2)

√(1 – z

^{2})dy = (dz/2)dy =

On integrating both sides, we get

∫dy = ∫

y = (1/2)sin

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

^{-1}(x^{2}) + c -(Here, ‘c’ is integration constant)

### Question 17. √(a + x)(dy) + xdx = 0

**Solution:**

We have,

√(a + x)(dy) + xdx = 0

dy = dx

Let, (x + a) = z

^{2}On differentiating both sides, we get

dx = 2zdz

(x + a) = z

^{2}x = z

^{2 }– ady = -2[(z

^{2 }– a)/z]zdzOn integrating both sides, we get

∫dy = -2∫[(z

^{2 }– a)/z]zdzy = -(2/3)(z

^{3}) + 2az + cy = -(2/3)(x + a)

^{3/2 }+ 2a√(x + a) + c -(Here, ‘c’ is integration constant)

### Question 18. (1 + x^{2})(dy/dx) – x = 2tan^{-1}x

**Solution:**

We have,

(1 + x^{2})(dy/dx) – x = 2tan^{-1}x

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

dy/dx =

dy

On integrating both sides, we get

y = I_{1 }+ I_{2}

I_{1 }= ∫(

Let, tan^{-1}x = z

On differentiating both sides, we get

= dz

= ∫2zdx

= z^{2}

I_{1 }= (tan^{-1}x)^{2}

I_{2 }= ∫

= (1/2)log|1 + x^{2}|

y = (tan^{-1}x)^{2 }+ 1/2log|1 + x^{2}|+ c -(Here, ‘c’ is integration constant)

### Question 19. (dy/dx) = xlogx

**Solution:**

We have,

(dy/dx) = xlogx

dy = xlogxdx

On integrating both sides, we get

∫dy = ∫xlogxdx

y = log|x|∫xdx – ∫[∫xdx]dx

y = (x

^{2}/2)log|x| – ∫(1/x)(x^{2}/2)dxy = (x

^{2}/2)log|x| – ∫(x/2)dxy = (x

^{2}/2)log|x| – (x^{2}/4) + c -(Here, ‘c’ is integration constant)

### Question 20. (dy/dx) = xe^{x }– (5/2) + cos^{2}x

**Solution:**

We have,

(dy/dx) = xe

^{x }– (5/2) + cos^{2}xdy = (xe

^{x }– (5/2) + cos^{2}x) dxOn integrating both sides, we get

∫dy = ∫xe

^{x }dx – 5/2∫dx + ∫cos^{2}x dxy = ∫xe

^{x }dx – 5/2∫dx + ∫(1 + cos2x)/2 dx= ∫xe

^{x }dx – 5/2∫dx + 1/2∫dx + 1/2∫cos2x dx= ∫xe

^{x }dx – 2∫dx + 1/2∫cos2x dx= x∫e

^{x }dx – ∫(1∫e^{x }dx)dx – 2x + sin2x/4 dx= xe

^{x }– e^{x }– 2x + 1/4sin2x + c

### Question 21. (x^{3 }+ x^{2 }+ x + 1)(dy/dx) = 2x^{2 }+ x

**Solution:**

We have,

(x^{3 }+ x^{2 }+ x + 1)(dy/dx) = 2x^{2 }+ x

(dy/dx) = (2x^{2 }+ x)/(x^{3 }+ x^{2 }+ x + 1)

dy =

On integrating both sides, we get

∫dy = ∫

Let,

2x^{2 }+ x = Ax^{2 }+ A + Bx^{2 }+ Bx + Cx + C

2x^{2 }+ x = (A + B)x^{2 }+ (B + C)x + (A + C)

On comparing the coefficients on both sides,

(A + B) = 2

(B + C) = 1

(A + C) = 0

After solving the equations,

A = (1/2)

B = (3/2)

C = -(1/2)

y = (1/2)∫(dx/(x + 1) +

y = (1/2)log(x + 1) + (3/4)∫dx – (1/2)∫

y = (1/2)log|x + 1| + (3/4)log|x^{2 }+ 1| – (1/2)tan^{-1}x + c -(Here, ‘c’ is integration constant)

### Question 22. sin(dy/dx) = k, y(0) = 1

**Solution:**

We have,

sin(dy/dx) = k,

(dy/dx) = sin

^{-1}(k)dy = sin

^{-1}(k)dxOn integrating both sides, we get

∫dy = sin

^{-1}(k)∫dxy = xsin

^{-1}k + c -(1)Put x = 0, y = 1

1 = 0 + c

1 = c

On putting the value of c in equation(1)

y = xsin

^{-1}k + 1y – 1 = xsin

^{-1}x

### Question 23. e^{(dy/dx) }= x + 1, y(0) = 3

**Solution:**

We have,

e

^{(dy/dx) }= x + 1(dy/dx) = log(x + 1)

dy = log(x + 1)dx

On integrating both sides, we get

∫dy = ∫log(x + 1)dx

y = log(x + 1)∫dx – ∫[∫dx]dx

y = xlog(x + 1) – ∫[x/(x + 1)]dx

y = xlog(x + 1) – ∫1 – \frac{1}{(x+1)}dx

y = xlog(x + 1) – x + log(x + 1) + c

y = (x + 1)log(x + 1) – x + c -(1)

Put, y = 3, x = 0 in equation(1)

3 = 0 + c

y = (x + 1)log(x + 1) – x + 3

### Question 24. c'(x) = 2 + 0.15x, c(0) = 100

**Solution:**

We have,

c'(x) = 2 + 0.15x -(1)

On integrating both sides, we get

∫c'(x)dx = ∫(2 + 0.15x)dx

c(x) = 2x + 0.15(x

^{2}/2) + c -(2)Put, c(0) = 100, x = 0 in equation(2)

100 = 2(0) + 0 + c

c = 100

c(x) = 2x + 0.15(x

^{2}/2) + 100

### Question 25. x(dy/dx) + 1 = 0, y(-1) = 0

**Solution:**

We have,

x(dy/dx) + 1 = 0

xdy = -dx

dy = -(dx/x)

On integrating both sides, we get

∫dy = -∫(dx/x)

y = -logx + c -(1)

Put, y = 0, x = -1 in equation(1)

0 = 0 + c

c = 0

y = -log|x|

### Question 26. x(x^{2 }– 1)(dy/dx) = 1, y(2) = 0

**Solution:**

We have,

x(x^{2 }– 1)(dy/dx) = 1 -(1)

dy = dx/x(x + 1)(x – 1)

On integrating both sides, we get

∫dy = ∫dx/x(x + 1)(x – 1)

Let, 1/x(x + 1)(x – 1) = A/x + B/(x + 1) + C/(x – 1)

1 = A(x + 1)(x – 1) + B(x)(x – 1) + C(x)(x + 1) -(2)

Put, x = 0, -1, 1 respectively and simplify above equation, we get,

A = -1, B = (1/2), C = (1/2)

y =

y = -logx + (1/2)log(x + 1) + (1/2)log(x – 1)

y = (1/2)log(1/x^{2}) + (1/2)log(x + 1) + (1/2)log(x – 1) + c -(3)

Put, y = 0, x = 2 in equation(3)

0 = (1/2)log(1/4) + (1/2)log(3) + 0 + c

c = -(1/2)log(3/4)

y = (1/2)log[(x^{2 }– 1)/x^{2}] – (1/2)log(3/4)

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.