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

Solve the following differential equations:

Question 1. (dy/dx) = x² + x – (1/x)

Solution:

We have,

(dy/dx) = x² + x – (1/x) 

dy = [x² + x – (1/x)]dx 

On integrating both sides, we get

∫(dy) = ∫[x² + 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) = x⁵ + x² – (2/x)     

On integrating both sides, we get

∫(dy) = ∫[x⁵ + x² – (2/x)]dx

y = (x⁶/6) + (x³/3) – 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]dx

On integrating both sides, we get

∫dy = ∫[-2x + e^3x]dx

y = (-2x²/2) + (e^3x/3) + c

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

Question 4. (x² + 1)(dy/dx) = 1

Solution:

We have,

(x² + 1)(dy/dx) = 1               

dy = (1/x² + 1)dx

On integrating both sides, we get

∫dy = ∫(dx/x² + 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) = x² + 3x + 7 

(dy/dx) = (x² + 3x + 7)/(x + 2)

(dy/dx) = (x² + 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^-1x 

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)(x² + 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 = I₁ + I₂

I₁ = 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

I₂ = x√(2x + 1)dx

Let, (2x + 1) = u²

On differentiating both sides 

2dx = 2udu

dx = udu

= [(u² – 1)/2]u × udu

I₂ = (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 = I₁ + 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 = I₁ + I₂

I₁ = (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

I₂ = 1/(xlogx)dx

Let, logx = z

On differentiating both sides, we get

(dx/x) = dz

= (dz/z)

= logz

= log(logx) + c₂

y = I₁ + 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, x³ = 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 ^{– 1}z ∫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)(z² + 1)tan^-1z – (z/6) + c     

y = (1/6)[(x⁶ + 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, x² = 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 = z² – a

dy = -2[(z² – a)/z]zdz

On integrating both sides, we get

∫dy = -2∫[(z² – 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 = I₁ + I₂

I₁ = ∫(

Let, tan^-1x = z

On differentiating both sides, we get 

 = dz

= ∫2zdx

= z²

I₁ = (tan^-1x)²

I₂ = ∫

= (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

∫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) = xe^x – (5/2) + cos²x

Solution:

We have,

(dy/dx) = xe^x – (5/2) + cos²x          

dy = (xe^x – (5/2) + cos²x) dx

On integrating both sides, we get

∫dy = ∫xe^x dx – 5/2∫dx + ∫cos²x dx

y = ∫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. (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

∫dy = ∫

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)∫(dx/(x + 1) + 

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

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

∫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(x² – 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[(x² – 1)/x²] – (1/2)log(3/4)

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