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

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

### Question 1. (dy/dx) = x2 + x – (1/x)

Solution:

We have,

(dy/dx) = x+ x – (1/x)

dy = [x+ x – (1/x)]dx

On integrating both sides, we get

∫(dy) = ∫[x+ x – (1/x)]dx

y = (x3/3) + (x2/2) – log(x) + c         -(Here, ‘c’ is integration constant)

### Question 2. (dy/dx) = x5 + x2 – (2/x)

Solution:

We have,

(dy/dx) = x+ x– (2/x)

On integrating both sides, we get

∫(dy) = ∫[x+ x– (2/x)]dx

y = (x6/6) + (x3/3) – 2log(x) + c         -(Here, ‘c’ is integration constant)

### Question 3. (dy/dx) + 2x = e3x

Solution:

We have,

(dy/dx) + 2x = e3x

(dy/dx) = -2x + e3x

dy = [-2x + e3x]dx

On integrating both sides, we get

∫dy = ∫[-2x + e3x]dx

y = (-2x2/2) + (e3x/3) + c

y = (-x2) + (e3x/3) + c         -(Here, ‘c’ is integration constant)

### Question 4. (x2 + 1)(dy/dx) = 1

Solution:

We have,

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

dy = (1/x+ 1)dx

On integrating both sides, we get

∫dy = ∫(dx/x+ 1)

y = tan-1x + 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) = (2sin2x/2)/(2cos2x/2)

dy/dx = tan2(x/2)

dy/dx = [sec2(x/2) – 1]

On integrating both sides, we get

∫dy = ∫[sec2(x/2) – 1]dx

y = [ tan(x/2)] – x + c

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

### Question 6. (x + 2)(dy/dx) = x2 + 3x + 7

Solution:

We have,

(x + 2)(dy/dx) = x+ 3x + 7

(dy/dx) = (x+ 3x + 7)/(x + 2)

(dy/dx) = (x+ 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 = x2/2 + x + 5log(x + 2) + c         -(Here, ‘c’ is integration constant)

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

Solution:

We have,

(dy/dx) = tan-1x

dy = tan-1xdx

On integrating both sides, we get

∫dy = ∫tan-1xdx

y = ∫1 × tan-1xdx

y = tan-1x∫1dx – ∫[ ∫1dx]dx

y = xtan-1x – ∫\frac{x}{(1 + x2)}dx

y = xtan-1x – y = xtan-1x – (1/2)log(1 + x2) + 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-1x

Solution:

We have,

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

dy = xtan-1x dx

On integrating both sides, we get

∫dy = ∫xtan-1x dx

y = tan-1x∫xdx – ∫[ ∫xdx]dx

y = (x2/2)tan-1x – y = (x2/2)tan-1x – (1/2)∫1- \frac{1}{(1 + x^2)}dx

y = (x2/2)tan-1x – (x/2) + (1/2)tan-1x + c

y = (1/2)(x+ 1)tan-1x – (x/2) + c         -(Here, ‘c’ is integration constant)

### Question 10. (dy/dx) = cos3xsin2x + x√(2x + 1)

Solution:

We have,

(dy/dx) = cos3xsin2x + x√(2x + 1)

(dy/dx) = cosxcos2xsin2x + x√(2x + 1)

y = I+ I2

I= cosxcos2xsin2xdx

On integrating both sides, we get

= ∫(1 – sin2x)sin2xcosxdx

Let, sinx = z

On differentiating both sides

cosxdx = dz

= ∫(1 – z2)z2dz

= z3/3 – z5/5 + c1

= sin3x/3 – sin5x/5 + c1

I= x√(2x + 1)dx

Let, (2x + 1) = u2

On differentiating both sides

2dx = 2udu

dx = udu

= [(u– 1)/2]u × udu

I= (1/2)(u4-u2)du

On integrating both sides, we get

= (1/2)[u5/5 – u3/3] + c2

= 1/10(2x + 1)5/2 – 1/6(2x + 1)3/2 + c2

y = I+ I2

y = (sin3x/3) – (sin5x/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) – xsin2x = 1/(xlogx)

Solution:

We have,

(dy/dx) – xsin2x = 1/(xlogx)

dy = xsin2xdx + 1/(xlogx)dx

On integrating both sides, we get

y = ∫xsin2xdx + ∫dx/(xlogx)

y = I+ I2

I= (1/2)(2sin2x)xdx

= (1/2)[(1 – cos2x)xdx]

= (1/2)(xdx – xcos2x)dx

On integrating both sides, we get

= 1/2[∫xdx – ∫xcos2x dx]

= 1/2(x2/2) – 1/2[x∫cos2x dx – ∫(1∫cos2x dx)]dx

= 1/2(x2/2) – (x/4)sin2x + ∫(1/4sin2x)dx

= 1/2(x2/2) – (x/4)sin2x + ((1/8)cos2x) + c1

I= 1/(xlogx)dx

Let, logx = z

On differentiating both sides, we get

(dx/x) = dz

= (dz/z)

= logz

= log(logx) + c2

y = I+ I2

y = (x2/4) – (xsin2x/4) + (cos2x/8) + log(logx) + c

### Question 13. (dy/dx) = x5tan-1(x3)

Solution:

We have,

(dy/dx) = x5tan-1(x3)

Let, x= z

On differentiating both sides, we get

3x2dx = dz

x2dx = dz/3

dy = (1/3)[ztan-1z]dz

On integrating both sides, we get

∫dy = (1/3)∫ztan-1z dz

y = (1/3)[tan – 1z ∫zdz – ∫{ ∫zdz}dz]

y = (z2/6)tan-1z – y = (z2/6)tan-1z – (1/6)∫1 – \frac{1}{(1 + z^2)}dz

y = (z2/6)tan-1z – (1/6)∫dz – (1/6)∫dz/(1 + z2)

y = (z2/6)tan-1z – (z/6) – (1/6)tan-1z

y = (1/6)(z+ 1)tan-1z – (z/6) + c

y = (1/6)[(x+ 1)tan – 1(x3) – (x3)] + c

### 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

∫(dy) = ∫(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

∫dy = ∫[2cos2x – 1 + 2cosx – secx]dx

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

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

Solution:

We have,

√(1 – x4)(dy/dx) = xdx

Let, x= z

On differentiating both sides, we get

2xdx = dz

xdx = (dz/2)

√(1 – z2)dy = (dz/2)

dy = On integrating both sides, we get

∫dy = ∫ y = (1/2)sin-1(z) + c

y = (1/2)sin-1(x2) + 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) = z2

On differentiating both sides, we get

dx = 2zdz

(x + a) = z2

x = z– a

dy = -2[(z– a)/z]zdz

On integrating both sides, we get

∫dy = -2∫[(z– a)/z]zdz

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

y = -(2/3)(x + a)3/2 + 2a√(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 = I+ I2

I= ∫( Let, tan-1x = z

On differentiating both sides, we get = dz

= ∫2zdx

= z2

I= (tan-1x)2

I= ∫ = (1/2)log|1 + x2|

y = (tan-1x)+ 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

∫dy = ∫xlogxdx

y = log|x|∫xdx – ∫[ ∫xdx]dx

y = (x2/2)log|x| – ∫(1/x)(x2/2)dx

y = (x2/2)log|x| – ∫(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) = xe– (5/2) + cos2x

dy = (xe– (5/2) + cos2x) dx

On integrating both sides, we get

∫dy = ∫xedx – 5/2∫dx + ∫cos2x dx

y = ∫xedx – 5/2∫dx + ∫(1 + cos2x)/2 dx

= ∫xedx – 5/2∫dx + 1/2∫dx + 1/2∫cos2x dx

= ∫xedx – 2∫dx + 1/2∫cos2x dx

= x∫edx – ∫(1∫edx)dx – 2x + sin2x/4 dx

= xe– e– 2x + 1/4sin2x + c

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

Solution:

We have,

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

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

dy = On integrating both sides, we get

∫dy = ∫ Let,

2x+ x = Ax+ A + Bx+ Bx + Cx + C

2x+ x = (A + B)x+ (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+ 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

∫dy = sin-1(k)∫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

∫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(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

∫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(x2 – 1)(dy/dx) = 1, y(2) = 0

Solution:

We have,

x(x– 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/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[(x– 1)/x2] – (1/2)log(3/4)

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