# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.5 | Set 2

Last Updated : 21 Feb, 2021

### Question 14. sin4x(dy/dx) = cosx

Solution:

We have,

sin4x(dy/dx) = cosx

dy = (cosx/sin4x)dx

Let, sinx = z

On differentiating both sides, we get

cosx dx = dz

dy = (dz/z4)

On integrating both sides, we get

âˆ«(dy) = âˆ«(1/z4)dz

y = (1/ -3t3) + c

y = -(1/3sin3x) + c

y = (-cosec3x/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) = (4cos3x – 3cosx + 2cos2x – 1)/cosx

(dy/dx) = (4cos3x/cosx) – 3(cosx/cosx) + 2(cos2x/cosx) – secx

dy = [4cos2x – 3 + 2cosx – secx]dx

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

On integrating both sides, we get

âˆ«dy = âˆ«[2cos2x – 1 + 2cosx – secx]dx

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

### Question 16. âˆš(1 – x4)(dy/dx) = xdx

Solution:

We have,

âˆš(1 – x4)(dy/dx) = xdx

Let, x2 = z

On differentiating both sides, we get

2xdx = dz

xdx = (dz/2)

âˆš(1 – z2)dy = (dz/2)

dy =

On integrating both sides, we get

âˆ«dy = âˆ«

y = (1/2)sin-1(z) + c

y = (1/2)sin-1(x2) + c          -(Here, ‘c’ is integration constant)

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

Solution:

We have,

âˆš(a + x)(dy) + xdx = 0

dy = dx

Let, (x + a) = z2

On differentiating both sides, we get

dx = 2zdz

(x + a) = z2

x = z2 – a

dy = -2[(z2 – a)/z]zdz

On integrating both sides, we get

âˆ«dy = -2âˆ«[(z2 – a)/z]zdz

y = -(2/3)(z3) + 2az + c

y = -(2/3)(x + a)3/2 + 2aâˆš(x + a) + c          -(Here, ‘c’ is integration constant)

### Question 18. (1 + x2)(dy/dx) – x = 2tan-1x

Solution:

We have,

(1 + x2)(dy/dx) – x = 2tan-1x

(1 + x2)(dy/dx) = 2tan-1x + x

dy/dx =

dy

On integrating both sides, we get

y = I1 + I2

I1 = âˆ«(

Let, tan-1x = z

On differentiating both sides, we get

= dz

= âˆ«2zdx

= z2

I1 = (tan-1x)2

I2 = âˆ«

= (1/2)log|1 + x2|

y = (tan-1x)2 + 1/2log|1 + x2|+ 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

âˆ«dy = âˆ«xlogxdx

y = log|x|âˆ«xdx – âˆ«[âˆ«xdx]dx

y = (x2/2)log|x| – âˆ«(1/x)(x2/2)dx

y = (x2/2)log|x| – âˆ«(x/2)dx

y = (x2/2)log|x| – (x2/4) + c          -(Here, ‘c’ is integration constant)

### Question 20. (dy/dx) = xex – (5/2) + cos2x

Solution:

We have,

(dy/dx) = xex – (5/2) + cos2x

dy = (xex – (5/2) + cos2x) dx

On integrating both sides, we get

âˆ«dy = âˆ«xex dx – 5/2âˆ«dx + âˆ«cos2x dx

y = âˆ«xex dx – 5/2âˆ«dx + âˆ«(1 + cos2x)/2 dx

= âˆ«xex dx – 5/2âˆ«dx + 1/2âˆ«dx + 1/2âˆ«cos2x dx

= âˆ«xex dx – 2âˆ«dx + 1/2âˆ«cos2x dx

= xâˆ«ex dx – âˆ«(1âˆ«ex dx)dx – 2x + sin2x/4 dx

= xex – ex – 2x + 1/4sin2x + c

### Question 21. (x3 + x2 + x + 1)(dy/dx) = 2x2 + x

Solution:

We have,

(x3 + x2 + x + 1)(dy/dx) = 2x2 + x

(dy/dx) = (2x2 + x)/(x3 + x2 + x + 1)

dy =

On integrating both sides, we get

âˆ«dy = âˆ«

Let,

2x2 + x = Ax2 + A + Bx2 + Bx + Cx + C

2x2 + x = (A + B)x2 + (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)âˆ«(dx/(x + 1) +

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

y = (1/2)log|x + 1| + (3/4)log|x2 + 1| – (1/2)tan-1x + 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)dx

On integrating both sides, we get

âˆ«dy = sin-1(k)âˆ«dx

y = xsin-1k + c              -(1)

Put x = 0, y = 1

1 = 0 + c

1 = c

On putting the value of c in equation(1)

y = xsin-1k + 1

y – 1 = xsin-1x

### 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

âˆ«dy = âˆ«log(x + 1)dx

y = log(x + 1)âˆ«dx – âˆ«[âˆ«dx]dx

y = xlog(x + 1) – âˆ«[x/(x + 1)]dx

y = xlog(x + 1) – âˆ«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

âˆ«c'(x)dx = âˆ«(2 + 0.15x)dx

c(x) = 2x + 0.15(x2/2) + c               -(2)

Put, c(0) = 100, x = 0 in equation(2)

100 = 2(0) + 0 + c

c = 100

c(x) = 2x + 0.15(x2/2) + 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

âˆ«dy = -âˆ«(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(x2 – 1)(dy/dx) = 1, y(2) = 0

Solution:

We have,

x(x2 – 1)(dy/dx) = 1             -(1)

dy = dx/x(x + 1)(x – 1)

On integrating both sides, we get

âˆ«dy = âˆ«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/x2) + (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[(x2 – 1)/x2] – (1/2)log(3/4)

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