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# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.5 | Set 1

• Last Updated : 21 Feb, 2021

### 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

âˆ«(dy) = âˆ«[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

âˆ«(dy) = âˆ«[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

âˆ«dy = âˆ«[-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

âˆ«dy = âˆ«(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

âˆ«dy = âˆ«[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

âˆ«dy = âˆ«(x + 1)dx + 5âˆ«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

âˆ«dy = âˆ«tan-1xdx

y = âˆ«1 Ã— tan-1xdx

y = tan-1xâˆ«1dx – âˆ«[âˆ«1dx]dx

y = xtan-1x – âˆ«\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

âˆ«dy = logxâˆ«1dx – âˆ«[âˆ«1dx]dx

y = xlogx – âˆ«[x/x]dx

y = xlogx – âˆ«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

âˆ«dy = âˆ«xtan-1x dx

y = tan-1xâˆ«xdx – âˆ«[âˆ«xdx]dx

y = (x2/2)tan-1x –

y = (x2/2)tan-1x – (1/2)âˆ«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 + xâˆš(2x + 1)

Solution:

We have,

(dy/dx) = cos3xsin2x + xâˆš(2x + 1)

(dy/dx) = cosxcos2xsin2x + xâˆš(2x + 1)

y = I1 + I2

I1 = cosxcos2xsin2xdx

On integrating both sides, we get

= âˆ«(1 – sin2x)sin2xcosxdx

Let, sinx = z

On differentiating both sides

cosxdx = dz

= âˆ«(1 – z2)z2dz

= z3/3 – z5/5 + c1

= sin3x/3 – sin5x/5 + c1

I2 = xâˆš(2x + 1)dx

Let, (2x + 1) = u2

On differentiating both sides

2dx = 2udu

dx = udu

= [(u2 – 1)/2]u Ã— 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

âˆ«dy = âˆ«(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 = âˆ«xsin2xdx + âˆ«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[âˆ«xdx – âˆ«xcos2x dx]

= 1/2(x2/2) – 1/2[xâˆ«cos2x dx – âˆ«(1âˆ«cos2x dx)]dx

= 1/2(x2/2) – (x/4)sin2x + âˆ«(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

âˆ«dy = (1/3)âˆ«ztan-1z dz

y = (1/3)[tan – 1z âˆ«zdz – âˆ«{âˆ«zdz}dz]

y = (z2/6)tan-1z –

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

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