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

### 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

âˆ«(dy) = âˆ«(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

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 – x^{4})(dy/dx) = xdx

**Solution:**

We have,

âˆš(1 – x

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

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

2xdx = dz

xdx = (dz/2)

âˆš(1 – z

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

On integrating both sides, we get

âˆ«dy = âˆ«

y = (1/2)sin

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

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

âˆ«dy = -2âˆ«[(z

^{2 }– a)/z]zdzy = -(2/3)(z

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

^{3/2 }+ 2aâˆš(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 + xdy/dx =

dy

On integrating both sides, we get

y = I

_{1 }+ I_{2}I

_{1 }= âˆ«(Let, tan

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

= dz

= âˆ«2zdx

= z

^{2}I

_{1 }= (tan^{-1}x)^{2}I

_{2 }= âˆ«= (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

âˆ«dy = âˆ«xlogxdx

y = log|x|âˆ«xdx – âˆ«[âˆ«xdx]dx

y = (x

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

^{2}/2)log|x| – âˆ«(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

âˆ«dy = âˆ«xe

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

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

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

^{x }dx – 2âˆ«dx + 1/2âˆ«cos2x dx= xâˆ«e

^{x }dx – âˆ«(1âˆ«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

âˆ«dy = âˆ«

Let,

2x

^{2 }+ x = Ax^{2 }+ A + Bx^{2 }+ Bx + Cx + C2x

^{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)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|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

âˆ«dy = sin

^{-1}(k)âˆ«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

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(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

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(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

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/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)

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