# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.9 | Set 2

### Question 14. 3x^{2}dy = (3xy + y^{2})dx

**Solution:**

We have,

3x

^{2}dy = (3xy + y^{2})dx(dy/dx) = (3xy + y

^{2})/3x^{2}It is a homogeneous equation,

So put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (3xvx + v

^{2}x^{2})/3x^{2}v + x(dv/dx) = (3v + v

^{2})/3x(dv/dx) = [(3v + v

^{2})/3] – vx(dv/dx) = (3v + v

^{2 }– 3v)/33(dv/v

^{2}) = (dx/x)On integrating both sides,

3∫(dv/v

^{2}) = ∫(dx/x)-(3/v) = log|x| + c

-3x/y = log(x) + c (Where ‘c’ is integration constant)

### Question 15. (dy/dx) = x/(2y + x)

**Solution:**

We have,

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

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = x/(2vx + x)

v + x(dv/dx) = 1/(2v + 1)

x(dv/dx) = [1/(2v + 1)] – v

x(dv/dx) = (1 – 2v

^{2 }– v)/(2v + 1)(2v + 1)dv/(2v

^{2 }+ v – 1) = -(dx/x)On integrating both sides,

∫(2v + 1)dv/(2v

^{2 }+ v – 1) = -∫(dx/x)Solving by partial fraction,

A(v + 1) + B(2v – 1) = (2v + 1) (i)

Putting v = -1 and solve above equation,

A(0) + B(-3) = (-1)

B = (1/3)

Putting v = -(1/2) and solve equation (i),

A(3/2) + B(0) = 2

A = (4/3)

(3/2)log|2v – 1| + (1/3)log|v + 1| = -log|x| + log|c|

log|(2v – 1)

^{2}(v + 1)| = -log|x|^{3 }+ log|c||(2v – 1)

^{2}(v + 1)| = (c/x^{3})(2y/x – 1)

^{2}(y/x + 1) = (c/x^{3})(2y – x)

^{2}(x + y) = c (Where ‘c’ is integration constant)

### Question 16. (x + 2y)dx – (2x – y)dy = 0

**Solution:**

We have,

(x + 2y)dx – (2x – y)dy = 0

(dy/dx) = (x + 2y)/(2x – y)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (x + 2vx)/(2x – vx)

v + x(dv/dx) = (1 + 2v)/(2 – v)

x(dv/dx) = [(1 + 2v)/(2 – v)] – v

x(dv/dx) = (1 + 2v – 2v + v

^{2})/(2 – v)(2 – v)dv/(1 + v

^{2}) = (dx/x)On integrating both sides,

∫(2 – v)dv/(1 + v

^{2}) = ∫(dx/x)2∫dv/(1 + v

^{2}) – ∫vdv/(1 + v^{2}) = log|x| + log|c|2tan

^{-1}v – (1/2)∫2vdv/(1 + v^{2}) = log|x| + log|c|2tan

^{-1}v – log|1 + v^{2}|^{1/2 }= log|cx|2tan

^{-1}v = log|cx√(1 + v^{2})|(Where ‘c’ is integration constant)

### Question 17.

**Solution:**

We have,

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (vx/x) – √(v

^{2}x^{2}/x^{2 }– 1)v + x(dv/dx) = v – √(v

^{2 }– 1)x(dv/dx) = -√(v

^{2 }– 1)dv/√(v

^{2 }– 1) = -(dx/x)On integrating both sides,

∫dv/√(v

^{2 }– 1) = -∫(dx/x)log|v + √(v

^{2 }– 1)| = -log|x| + log|c||v + √(v

^{2 }– 1)| = (c/x)y + √(y

^{2 }– x^{2}) = c (Where ‘c’ is integration constant)

### Question 18. (dy/dx) = (y/x){log(y) – log(x) + 1}

**Solution:**

We have,

(dy/dx) = (y/x){log(y) – log(x) + 1}

(dy/dx) = (y/x){log(y/x) + 1}

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v{log(v) + 1}

v + x(dv/dx) = vlog(v) + v

x(dv/dx) = vlog(v)

dv/vlogv = (dx/x)

On integrating both sides,

∫dv/vlogv = ∫(dx/x)

Let, logv = z

On differentiating both sides,dv/v = dz

∫(dz/z) = ∫(dx/x)

log|z| = log|x| + log|c|

z = xc

log|v| = xc

log|y/x| = xc (Where ‘c’ is integration constant)

### Question 19. (dy/dx) = (y/x) + sin(y/x)

**Solution:**

We have,

(dy/dx) = (y/x) + sin(y/x)

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v + sin(v)

x(dv/dx) = sin(v)

dv/sin(v) = (dx/x)

On integrating both sides,

∫dv/sin(v) = ∫(dx/x)

∫cosec(v)dv = ∫(dx/x)

log|tan(v/2)| = log(x) + log(c)

log|tan(y/2x)| = log|xc|

tan(y/2x) = |xc| (Where ‘c’ is integration constant)

### Question 20. y^{2}dx + (x^{2 }– xy + y^{2})dy = 0

**Solution:**

We have,

y

^{2}dx + (x^{2 }– xy + y^{2})dy = 0(dy/dx) = -(y

^{2})/(x^{2 }– xy + y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = -(v

^{2}x^{2})/(x^{2 }– xvx + v^{2}x^{2})v + x(dv/dx) = -(v

^{2})/(1 – v + v^{2})y(dv/dx) = [-(v

^{2})/(1 – v + v^{2})] – vOn integrating both sides,

∫dv/(1 + v

^{2}) – ∫dv/v = ∫(dx/x)tan

^{-1}(v) – log(v) = log(x) + log(c)tan

^{-1}(y/x) – log|y/x| = log(xc)tan

^{-1}(y/x) = log|(y/x)xc|tan

^{-1}(y/x) = log|yc|(Where ‘c’ is integration constant)

### Question 21. [x√(x^{2 }+ y^{2}) – y^{2}]dx + xydy = 0

**Solution:**

We have,

[x√(x

^{2 }+ y^{2}) – y^{2}]dx + xydy = 0dy/dx = -[x√(x

^{2 }+ y^{2}) – y^{2}]/xydy/dx = [y

^{2 }– x√(x^{2 }+ y^{2})]/xyIt is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = [v

^{2}x^{2 }– x√(x^{2 }+ v^{2}x^{2})]/xvxv + x(dv/dx) = [v

^{2 }– √(1 + v^{2})]/vx(dv/dx) = [v

^{2 }– √(1 + v^{2})]/v – vx(dv/dx) = -(√(1 + v

^{2})/vvdv/√(1 + v

^{2}) = -(dx/x)On integrating both sides,

∫vdv/√(1 + v

^{2}) = -∫(dx/x)(1/2)∫2vdv/√(1 + v

^{2}) = -∫(dx/x)Let, 1 + v

^{2 }= z

On differentiating both sides,2vdv = dz

(1/2)∫dz/√z = -∫(dx/x)

√z = -log|x| + log|c|

√(1 + v

^{2}) = log|c/x|√(x

^{2 }+ y^{2})/x = log|c/x|√(x

^{2 }+ y^{2}) = xlog|c/x| (Where ‘c’ is integration constant)

### Question 22. x(dy/dx) = y – xcos^{2}(y/x)

**Solution:**

We have,

x(dy/dx) = y – xcos

^{2}(y/x)(dy/dx) = y/x – cos

^{2}(y/x)It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = v – cos

^{2}(v)x(dv/dx) = -cos

^{2}(v)dv/cos

^{2}(v) = -(dx/x)On integrating both sides,

∫dv/cos

^{2}(v) = -∫(dx/x)∫sec

^{2}vdv = -∫(dx/x)tan(v) = -log|x| + log|c|

tan(y/x) = log|c/x| (Where ‘c’ is integration constant)

### Question 23. (y/x)cos(y/x)dx – {(x/y)sin(y/x) + cos(y/x)}dy = 0

**Solution:**

We have,

(y/x)cos(y/x)dx – {(x/y)sin(y/x) + cos(y/x)}dy = 0

It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.t x,dy/dx = v + x(dv/dx)

So,

x(dv/dx) = (v

^{2}cosv – vsinv – v^{2}cosv)/(sinv + vcosv)x(dv/dx) = -vsinv/(sinv + vcosv)

On integrating both sides,

∫[(sinv + vcosv)/vsinv]dv = -∫(dx/x)

∫(dv/v) + ∫(cotv)dv = -∫(dx/x)

log|v| + log|sinv| = -log|x| + log|c|

log|vsinv| = log|c/x|

(y/x)sin(y/x) = (c/x)

ysin(y/x) = c (Where ‘c’ is integration constant)

### Question 24. xylog(x/y)dx + {y^{2 }– x^{2}log(x/y)}dy = 0

**Solution:**

We have,

xylog(x/y)dx + {y

^{2 }– x^{2}log(x/y)}dy = 0It is a homogeneous equation,

So, put x = vy (i)

On differentiating both sides w.r.ty,dx/dy = v + y(dv/dy)

So,

v + y(dv/dy) = (v

^{2}logv – 1)/(vlogv)y(dv/dy) = [(v

^{2}logv – 1)/(vlogv)] – vy(dv/dy) = (v

^{2}logv – 1 – v^{2}logv)/vlogvy(dv/dy) = -(1/vogv)

vlogvdv = -(dy/y)

On integrating both sides,

∫vlogvdv = -∫(dy/y)

logv∫vdv – ∫{d/dv(logv)∫vdv}dv}dv = -∫(dy/y)

(v

^{2}/2)logv – (1/2)∫(1/v)(v^{2}/2)dv = -logy + logc(v

^{2}/2)logv – (1/2)∫vdv = -logy + logc(v

^{2}/2)logv – (v^{2}/4) + logy = log|c|(v

^{2}/2)[logv – 1/2] + logy = log|c|v

^{2}[logv – (1/2)] + logy^{ }= log|c|(x

^{2}/y^{2})[log(x/y) – (1/2)] + logy^{ }= log|c| (Where ‘c’ is integration constant)

### Question 25. (1 + e^{x/y})dx + e^{x/y}(1 – x/y)dy = 0

**Solution:**

We have,

(1 + e

^{x/y})dx + e^{x/y}(1 – x/y)dy = 0It is a homogeneous equation,

So, put x = vy (i)

On differentiating both sides w.r.ty,dx/dy = v + y(dv/dy)

So,

y(dv/dy) = -[e

^{v}(1 – v)/(1 + e^{v})] – vy(dv/dy) = (-e

^{v }+ ve^{v }– v – ve^{v})/(1 + e^{v})y(dv/dy) = -(v + ve

^{v})/(1 + e^{v})[(1 + e

^{v})/(v + e^{v})]dv = -(dy/y)On integrating both sides,

∫[(1 + e

^{v})/(v + e^{v})]dv = -∫(dy/y)log|(v + e

^{v})| = -log(y) + log(c)log|(v + e

^{v})| = log|c/y|(x/y) + e

^{x/y }= c/yx + ye

^{x/y }= c (Where ‘c’ is integration constant)

### Question 26. (x^{2 }+ y^{2})dy/dx = (8x^{2 }– 3xy + 2y^{2})

**Solution:**

We have,

(x

^{2 }+ y^{2})dy/dx = (8x^{2 }– 3xy + 2y^{2})(dy/dx) = (8x

^{2 }– 3xy + 2y^{2})/(x^{2 }+ y^{2})It is a homogeneous equation,

So, put y = vx (i)

On differentiating both sides w.r.tx,dy/dx = v + x(dv/dx)

So,

v + x(dv/dx) = (8x

^{2 }– 3xvx + 2v^{2}x^{2})/(x^{2 }+ v^{2}x^{2})v + x(dv/dx) = (8 – 3v + 2v

^{2})/(1 + v^{2})x(dv/dx) = [(8 – 3v + 2v

^{2})/(1 + v^{2})] – vx(dv/dx) = (8 – 4v + 2v

^{2 }– v^{3})/(1 + v^{2})(1 + v

^{2})dv/(8 – 4v + 2v^{2 }– v^{3}) = (dx/x)On integrating both sides,

Using partial fraction,

(1 + v

^{2}) = Av(2 – v) + B(2 – v) + C(4 + v^{2})(1 + v

^{2}) = 2Av – Av^{2 }+ 2B – Bv + 4C + Cv^{2}(1 + v

^{2}) = (C – A)v^{2 }+ (2A – B)v + (2B + 4C)Comparing the co-efficient of both sides,

(C – A) = 1

(2A – B) = 0

(2B + 4C) = 1

Solving above equations,

A = -(3/8)

B = -(3/4)

C = (5/8)

(Where ‘c’ is integration constant)