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

Question 14. 3x²dy = (3xy + y²)dx

Solution:

We have,

3x²dy = (3xy + y²)dx

(dy/dx) = (3xy + y²)/3x²

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 + v²x²)/3x²

v + x(dv/dx) = (3v + v²)/3

x(dv/dx) = [(3v + v²)/3] – v

x(dv/dx) = (3v + v² – 3v)/3

3(dv/v²) = (dx/x)

On integrating both sides,

3∫(dv/v²) = ∫(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 – 2v² – v)/(2v + 1)

(2v + 1)dv/(2v² + v – 1) = -(dx/x)

On integrating both sides,

∫(2v + 1)dv/(2v² + 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/x³)

(2y/x – 1)²(y/x + 1) = (c/x³)

(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 + v²)/(2 – v)

(2 – v)dv/(1 + v²) = (dx/x)

On integrating both sides,

∫(2 – v)dv/(1 + v²) = ∫(dx/x)

2∫dv/(1 + v²) – ∫vdv/(1 + v²) = log|x| + log|c|

2tan^-1v – (1/2)∫2vdv/(1 + v²) = log|x| + log|c|

2tan^-1v – log|1 + v²|^1/2 = log|cx|

2tan^-1v = log|cx√(1 + v²)|

  (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) – √(v²x²/x² – 1)

v + x(dv/dx) = v – √(v² – 1)

x(dv/dx) = -√(v² – 1)

dv/√(v² – 1) = -(dx/x)

On integrating both sides,

∫dv/√(v² – 1) = -∫(dx/x)

log|v + √(v² – 1)| = -log|x| + log|c|

|v + √(v² – 1)| = (c/x)

y + √(y² – x²) = 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. y²dx + (x² – xy + y²)dy = 0

Solution:

We have,

y²dx + (x² – xy + y²)dy = 0

(dy/dx) = -(y²)/(x² – xy + y²)

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) = -(v²x²)/(x² – xvx + v²x²)

v + x(dv/dx) = -(v²)/(1 – v + v²)

y(dv/dx) = [-(v²)/(1 – v + v²)] – v

On integrating both sides,

∫dv/(1 + v²) – ∫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√(x² + y²) – y²]dx + xydy = 0

Solution:

We have,

[x√(x² + y²) – y²]dx + xydy = 0

dy/dx = -[x√(x² + y²) – y²]/xy

dy/dx = [y² – x√(x² + y²)]/xy

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) = [v²x² – x√(x² + v²x²)]/xvx

v + x(dv/dx) = [v² – √(1 + v²)]/v

x(dv/dx) = [v² – √(1 + v²)]/v – v

x(dv/dx) = -(√(1 + v²)/v

vdv/√(1 + v²) = -(dx/x)

On integrating both sides,

∫vdv/√(1 + v²) = -∫(dx/x)

(1/2)∫2vdv/√(1 + v²) = -∫(dx/x)

Let, 1 + v² = z

On differentiating both sides,

2vdv = dz

(1/2)∫dz/√z = -∫(dx/x)

√z = -log|x| + log|c|

√(1 + v²) = log|c/x|

√(x² + y²)/x = log|c/x|

√(x² + y²) = xlog|c/x|    (Where ‘c’ is integration constant)

Question 22. x(dy/dx) = y – xcos²(y/x)

Solution:

We have,

x(dy/dx) = y – xcos²(y/x)

(dy/dx) = y/x – cos²(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 – cos²(v)

x(dv/dx) = -cos²(v)

dv/cos²(v) = -(dx/x)

On integrating both sides,

∫dv/cos²(v) = -∫(dx/x)

∫sec²vdv = -∫(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) = (v²cosv – vsinv – v²cosv)/(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 + {y² – x²log(x/y)}dy = 0

Solution:

We have,

xylog(x/y)dx + {y² – x²log(x/y)}dy = 0

It is a homogeneous equation,

So, put x = vy               (i)

On differentiating both sides w.r.t y,

dx/dy = v + y(dv/dy)

So,

v + y(dv/dy) = (v²logv – 1)/(vlogv)

y(dv/dy) = [(v²logv – 1)/(vlogv)] – v

y(dv/dy) = (v²logv – 1 – v²logv)/vlogv

y(dv/dy) = -(1/vogv)

vlogvdv = -(dy/y)

On integrating both sides,

∫vlogvdv = -∫(dy/y)

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

(v²/2)logv – (1/2)∫(1/v)(v²/2)dv = -logy + logc

(v²/2)logv – (1/2)∫vdv = -logy + logc

(v²/2)logv – (v²/4) + logy = log|c|

(v²/2)[logv – 1/2] + logy = log|c|

v²[logv – (1/2)] + logy= log|c|

(x²/y²)[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 = 0

It is a homogeneous equation,

So, put x = vy               (i)

On differentiating both sides w.r.t y,

dx/dy = v + y(dv/dy)

So,

y(dv/dy) = -[e^v(1 – v)/(1 + e^v)] – v

y(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/y

x + ye^x/y = c  (Where ‘c’ is integration constant)

Question 26. (x² + y²)dy/dx = (8x² – 3xy + 2y²)

Solution:

We have,

(x² + y²)dy/dx = (8x² – 3xy + 2y²)

(dy/dx) = (8x² – 3xy + 2y²)/(x² + y²)

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) = (8x² – 3xvx + 2v²x²)/(x² + v²x²)

v + x(dv/dx) = (8 – 3v + 2v²)/(1 + v²)

x(dv/dx) = [(8 – 3v + 2v²)/(1 + v²)] – v

x(dv/dx) = (8 – 4v + 2v² – v³)/(1 + v²)

(1 + v²)dv/(8 – 4v + 2v² – v³) = (dx/x)

On integrating both sides,

Using partial fraction,

(1 + v²) = Av(2 – v) + B(2 – v) + C(4 + v²)

(1 + v²) = 2Av – Av² + 2B – Bv + 4C + Cv²

(1 + v²) = (C – A)v² + (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)