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

### Solve the following differential equations:

### Question 1. (dy/dx) = x^{2 }+ x – (1/x)

**Solution:**

We have,

(dy/dx) = x

^{2 }+ x – (1/x)dy = [x

^{2 }+ x – (1/x)]dxOn integrating both sides, we get

∫(dy) = ∫[x

^{2 }+ x – (1/x)]dxy = (x

^{3}/3) + (x^{2}/2) – log(x) + c -(Here, ‘c’ is integration constant)

### Question 2. (dy/dx) = x^{5 }+ x^{2 }– (2/x)

**Solution:**

We have,

(dy/dx) = x

^{5 }+ x^{2 }– (2/x)On integrating both sides, we get

∫(dy) = ∫[x

^{5 }+ x^{2 }– (2/x)]dxy = (x

^{6}/6) + (x^{3}/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}]dxOn integrating both sides, we get

∫dy = ∫[-2x + e

^{3x}]dxy = (-2x

^{2}/2) + (e^{3x}/3) + cy = (-x

^{2}) + (e^{3x}/3) + c -(Here, ‘c’ is integration constant)

### Question 4. (x^{2 }+ 1)(dy/dx) = 1

**Solution:**

We have,

(x

^{2 }+ 1)(dy/dx) = 1dy = (1/x

^{2 }+ 1)dxOn integrating both sides, we get

∫dy = ∫(dx/x

^{2 }+ 1)y = tan

^{-1}x + 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) = (2sin

^{2}x/2)/(2cos^{2}x/2)dy/dx = tan

^{2}(x/2)dy/dx = [sec

^{2}(x/2) – 1]On integrating both sides, we get

∫dy = ∫[sec

^{2}(x/2) – 1]dxy = [tan(x/2)] – x + c

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

### Question 6. (x + 2)(dy/dx) = x^{2 }+ 3x + 7

**Solution:**

We have,

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

^{2 }+ 3x + 7(dy/dx) = (x

^{2 }+ 3x + 7)/(x + 2)(dy/dx) = (x

^{2 }+ 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 = x

^{2}/2 + x + 5log(x + 2) + c -(Here, ‘c’ is integration constant)

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

**Solution:**

We have,

(dy/dx) = tan

^{-1}xdy = tan

^{-1}xdxOn integrating both sides, we get

∫dy = ∫tan

^{-1}xdxy = ∫1 × tan

^{-1}xdxy = tan

^{-1}x∫1dx – ∫[∫1dx]dxy = xtan

^{-1}x – ∫\frac{x}{(1 + x2)}dxy = xtan

^{-1}x –y = xtan

^{-1}x – (1/2)log(1 + x^{2}) + 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^{-1}x

**Solution:**

We have,

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

^{-1}xdy = xtan

^{-1}x dxOn integrating both sides, we get

∫dy = ∫xtan

^{-1}x dxy = tan

^{-1}x∫xdx – ∫[∫xdx]dxy = (x

^{2}/2)tan^{-1}x –y = (x

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

^{2}/2)tan^{-1}x – (x/2) + (1/2)tan^{-1}x + cy = (1/2)(x

^{2 }+ 1)tan^{-1}x – (x/2) + c -(Here, ‘c’ is integration constant)

### Question 10. (dy/dx) = cos^{3}xsin^{2}x + x√(2x + 1)

**Solution:**

We have,

(dy/dx) = cos

^{3}xsin^{2}x + x√(2x + 1)(dy/dx) = cosxcos

^{2}xsin^{2}x + x√(2x + 1)y = I

_{1 }+ I_{2}I

_{1 }= cosxcos^{2}xsin^{2}xdxOn integrating both sides, we get

= ∫(1 – sin

^{2}x)sin^{2}xcosxdxLet, sinx = z

On differentiating both sides

cosxdx = dz

= ∫(1 – z

^{2})z^{2}dz= z

^{3}/3 – z^{5}/5 + c_{1}= sin

^{3}x/3 – sin^{5}x/5 + c_{1}I

_{2 }= x√(2x + 1)dxLet, (2x + 1) = u

^{2}On differentiating both sides

2dx = 2udu

dx = udu

= [(u

^{2 }– 1)/2]u × uduI

_{2 }= (1/2)(u^{4}-u^{2})duOn integrating both sides, we get

= (1/2)[u

^{5}/5 – u^{3}/3] + c_{2}= 1/10(2x + 1)

^{5/2}– 1/6(2x + 1)^{3/2 }+ c_{2}y = I

_{1 }+ I_{2}y = (sin

^{3}x/3) – (sin^{5}x/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) – xsin^{2}x = 1/(xlogx)

**Solution:**

We have,

(dy/dx) – xsin

^{2}x = 1/(xlogx)dy = xsin

^{2}xdx + 1/(xlogx)dxOn integrating both sides, we get

y = ∫xsin

^{2}xdx + ∫dx/(xlogx)y = I

_{1 }+ I_{2}I

_{1 }= (1/2)(2sin^{2}x)xdx= (1/2)[(1 – cos2x)xdx]

= (1/2)(xdx – xcos2x)dx

On integrating both sides, we get

= 1/2[∫xdx – ∫xcos2x dx]

= 1/2(x

^{2}/2) – 1/2[x∫cos2x dx – ∫(1∫cos2x dx)]dx= 1/2(x

^{2}/2) – (x/4)sin2x + ∫(1/4sin2x)dx= 1/2(x

^{2}/2) – (x/4)sin2x + ((1/8)cos2x) + c_{1}I

_{2 }= 1/(xlogx)dxLet, logx = z

On differentiating both sides, we get

(dx/x) = dz

= (dz/z)

= logz

= log(logx) + c

_{2}y = I

_{1 }+ I_{2}y = (x

^{2}/4) – (xsin2x/4) + (cos2x/8) + log(logx) + c

### Question 13. (dy/dx) = x^{5}tan^{-1}(x^{3})

**Solution:**

We have,

(dy/dx) = x

^{5}tan^{-1}(x^{3})Let, x

^{3 }= zOn differentiating both sides, we get

3x

^{2}dx = dzx

^{2}dx = dz/3dy = (1/3)[ztan

^{-1}z]dzOn integrating both sides, we get

∫dy = (1/3)∫ztan

^{-1}z dzy = (1/3)[tan

^{ – 1}z ∫zdz – ∫{∫zdz}dz]y = (z

^{2}/6)tan^{-1}z –y = (z

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

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

^{2}/6)tan^{-1}z – (z/6) – (1/6)tan^{-1}zy = (1/6)(z

^{2 }+ 1)tan^{-1}z – (z/6) + cy = (1/6)[(x

^{6 }+ 1)tan – 1(x^{3}) – (x^{3})] + c

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