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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

∫(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 = \frac{dz}{2\sqrt{1-z^2}}

On integrating both sides, we get

∫dy = ∫\frac{dz}{2\sqrt{1-z^2}}

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 = \frac{-x}{\sqrt{a + x}}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 = \frac{2tan^{-1}x + x}{1 + x^2}

dy =(\frac{2tan^{-1}x+x}{1+x^2})dx

On integrating both sides, we get

∫dy=∫(\frac{2tan^{-1}x+x}{1+x^2})dx

y = I1 + I2

I1 = ∫(\frac{2tan^{-1}x}{1 + x^2}

Let, tan-1x = z

On differentiating both sides, we get 

\frac{dx}{1 + x^2} = dz

= ∫2zdx

= z2

I1 = (tan-1x)2

I2 = ∫\frac{xdx}{(1+x^2)}

= (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 – ∫[\frac{(logx)}{dx}∫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) = xex – (5/2) + cos2x          

dy = (xex – (5/2) + cos2x) dx

On integrating both sides, we get

∫dy = ∫xex dx – 5/2∫dx + ∫cos2x dx

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

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

= ∫xex dx – 2∫dx + 1/2∫cos2x dx

= x∫ex dx – ∫(1∫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 = \frac{2x^2+x}{(x+1)(x^2+1)}dx

On integrating both sides, we get

∫dy = ∫\frac{2x^2+x}{(x+1)(x^2+1)}dx

Let,

\frac{2x^2+x}{(x+1)(x^2+1)}dx=\frac{A}{(x+1)}+\frac{Bx+C}{x^2+1}

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) + ∫\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}dx

y = (1/2)log(x + 1) + (3/4)∫\frac{2x}{x^2 + 1}dx – (1/2)∫\frac{dx}{(x^2 + 1)}

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 – ∫[\frac{dlog(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 = -∫\frac{dx}{x}+∫\frac{1}{2}\frac{dx}{(x+1)}+∫\frac{1}{2}\frac{dx}{(x-1)}

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)

y=\frac{1}{2}log[\frac{4}{3}\frac{(x^2-1)}{x^2}]

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