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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.5 | Set 1
  • Last Updated : 21 Feb, 2021

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 = 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,

(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 = [\frac{1}{\frac{1}{2}}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 – ∫[\frac{d(tan^{-1}x)}{dx}∫1dx]dx

y = xtan-1x – ∫\frac{x}{(1 + x2)}dx

y = xtan-1x – \frac{1}{2}∫\frac{2x}{1+x^2}dx

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 – ∫[\frac{d(logx)}{dx}∫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 – ∫[\frac{d(tan^{-1}x)}{dx}∫xdx]dx

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

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 – 1z ∫zdz – ∫{\frac{d(tan ^{-1}z)}{dz}∫zdz}dz]

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

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             

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