RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	22
Exercise	22.9
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations

Solve the following differential equations:

Question 1. x²dy + y(x + y)dy = 0

Solution:

We have,

x²dy + y(x + y)dy = 0

dy/dx = -y(x + y)/x²

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 + vx)/x²

v + x(dv/dx) = -v – v²

x(dv/dx) = -2v – v²

$\frac{dv}{(v^2 + 2v)} = -\frac{dx}{x}$

On integrating both sides,

$∫\frac{dv}{(v^2+2v+1-1)}=-∫(\frac{dx}{x})$

$∫\frac{dv}{(v+1)^2-(1)^2}=-log(x)$

$\frac{1}{2}log(\frac{v+1-1}{v+1+1})=-log|x|+log|c|$

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

v/(v + 2) = c²/x²

$\frac{\frac{y}{x}}{\frac{y}{x}+2}=\frac{c^2}{x^2}$

yx²= (y + 2x)c² (Where ‘c’ is integration constant)

Question 2. (dy/dx) = (y – x)/(y + x)

Solution:

We have,

(dy/dx) = (y – x)/(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) = (vx – x)/(vx + x)

v + x(dv/dx) = (v – 1)/(v + 1)

x(dv/dx) = (v – 1)/(v + 1) – v

x(dv/dx) = (v – 1 – v²– v)/(v + 1)

x(dv/dx) = -(v²+ 1)/(v + 1)

$\frac{(v+1)dv}{(v^2+1)}=-\frac{dx}{x}$

On integrating both sides,

$∫\frac{(v+1)dv}{(v^2+1)}=-∫\frac{dx}{x}$

∫vdv/(v²+1)+∫dv/(v²+1)=-∫(dx/x)

$\frac{1}{2}∫\frac{2v}{v^2+1}+∫\frac{dv}{v^2+1}=-log|x|+log|c|$

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

log|(y²+ x²)/x²| + 2tan^-1(y/x) = log(c/x)²

log(y²+ x²) – log(x)²+ 2tan^-1(y/x) = log(c/x)²

log(y²+ x²) + 2tan^-1(y/x) = 2log(c) (Where ‘c’ is integration constant)

Question 3. (dy/dx) = (y²– x²)/2yx

Solution:

We have,

(dy/dx) = (y²– x²)/2yx

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)/2vx²

v + x(dv/dx) = (v²– 1)/2v

x(dv/dx) = [(v²– 1)/2v] – v

x(dv/dx) = (v²– 1 – 2v²)/2v

x(dv/dx) = -(v²+ 1)/2v

$\frac{2vdv}{(v^2 + 1)} = -\frac{dx}{x}$

On integrating both sides,

$∫\frac{vdv}{(v^2+1)}=-∫\frac{dx}{x}$

log|v²+1| = -log(x) + log(c)

log|v²+1| = log(c/x)

y²/x²+ 1 = |c/x|

(x²+ y²) = cx (Where ‘c’ is integration constant)

Question 4. x(dy/dx) = (x + y)

Solution:

We have,
x(dy/dx) = (x+y)
(dy/dx) = (x+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) = (x + vx)/x
v + x(dv/dx) = (1 + v)
x(dv/dx) = 1
dv = (dx/x)
On integrating both sides,
∫dv = ∫(dx/x)
v = log(x) + c
y/x = log(x) + c
y = xlog(x) + cx (Where ‘c’ is integration constant)

Question 5. (x²– y²)dx – 2xydy = 0

Solution:

We have,

(x²– y²)dx – 2xydy = 0

(dy/dx) = (x²– y²)/2xy

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) = (x²– v²x²)/2xvx

v + x(dv/dx) = (1 – v²)/2v

x(dv/dx) = [(1 – v²)/2v] – v

x(dv/dx) = (1 – 3v²)/2v

$\frac{2vdv}{(1-3v^2)} = \frac{dx}{x}$

On integrating both sides,

$∫\frac{2vdv}{(1-3v^2)}=∫\frac{dx}{x}$

$\frac{1}{3}∫\frac{6v}{1-3v^2}dv=∫\frac{dx}{x}$

-(1/3)log(1 – 3v²) = log(x) – log(c)

log(1 – 3v²) = -log(x)³+ log(c)

$(1-\frac{3y^2}{x^2})=(\frac{c}{x^3})$

(x²– 3y²)/x²= (c/x³)

x(x²– 3y²) = c (Where ‘c’ is integration constant)

Question 6. (dy/dx) = (x + y)/(x – y)

Solution:

We have,

(dy/dx) = (x + y)/(x – 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 + vx)/(x – vx)

v + x(dv/dx) = (1 + v)/(1 – v)

x(dv/dx) = [(1 + v)/(1 – v)] – v

x(dv/dx) = (1 + v – v + v²)/(1 – v)

x(dv/dx) = (1 + v²)/(1 – v)

$\frac{(1-v)dv}{(v^2+1)}=\frac{dx}{x}$

On integrating both sides,

$∫\frac{(v+1)dv}{(v^2+1)}=∫\frac{dx}{x}$

∫dv/(v²+ 1) – ∫vdv/(v²+ 1) = ∫(dx/x)

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

tan^-1(y/x) – (1/2)log(y²/x²+ 1) = log(x) + c

tan^-1(y/x) – (1/2)log(y²+ x²) + log(x) = log(x) + c

tan^-1(y/x) = (1/2)log(y²+ x²) + c (Where ‘c’ is integration constant)

Question 7. 2xy(dy/dx) = (x²+ y²)

Solution:

We have,

2xy(dy/dx) = (x²+ y²)

(dy/dx) = (x²+ y²)/2xy

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) = (x²+ v²x²)/2xvx

v + x(dv/dx) = (1 + v²)/2v

x(dv/dx) = [(1 + v²)/2v] – v

x(dv/dx) = (1 – v²)/2v

$\frac{2vdv}{(1 - v^2)} = \frac{dx}{x}$

On integrating both sides,

$∫\frac{2vdv}{(1 - v^2)} = ∫\frac{dx}{x}$

-log(1 – v²) = log(x) – log(c)

log(1 – v²) = -log(x) + log(c)

1 – y²/x²= (c/x)

(x²– y²) = cx (Where ‘c’ is integration constant)

Question 8. x²(dy/dx) = x²– 2y²+ xy

Solution:

We have,

x²(dy/dx) = x²– 2y²+ xy

(dy/dx) = (x²– 2y²+ xy)/x²

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) = (x²– 2v²x²+ xvx)/2xvx

v + x(dv/dx) = (1 – 2v²+ v)/x²

x(dv/dx) = (1 – 2v²+ v) – v

x(dv/dx) = (1 – 2v²)

dv/(1 – 2v²) = (dx/x)

On integrating both sides,

dv/(1 – 2v²) = ∫(dx/x)

$∫\frac{dv}{v^2-\frac{1}{2}}=-2\frac{dx}{x}$

$∫\frac{dv}{(\frac{1}{\sqrt2})^2-v^2}=2\frac{dx}{x}$

$\frac{1}{\sqrt{2}}log|\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}|=2log(x)+log(c)$

$\frac{1}{\sqrt{2}}log|\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}|=2log(x)+log(c)$

$(\frac{1}{√2})log|\frac{(x + y√2)}{(x - y√2)}|=log(x)^2 + log(c)$

$|\frac{(x + y√2)}{(x - y√2)}|^{\frac{1}{√2}} = cx^2$

$|\frac{(x+y√2)}{(x-y√2)}|=|cx^2|^{√2}$ (Where ‘c’ is integration constant)

Question 9. xy(dy/dx) = x²– y²

Solution:

We have,
xy(dy/dx) = x²– y²
(dy/dx) = (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) = (x²– v²x²)/xvx
v + x(dv/dx) = (1 – v²)/v
x(dv/dx) = [(1 – v²)/v] – v
x(dv/dx) = (1 – 2v²)/v
vdv/(1 – 2v²) = (dx/x)
On integrating both sides,
∫vdv/(1 – 2v²) = ∫(dx/x)
∫4vdv/(1 – 2v²) = 4∫(dx/x)
-log(1 – 2v²) = 4log(x) – log(c)
log(1 – 2v²) = log(c/x⁴)
(1 – 2y²/x²) = c/x⁴
(x²-2y²)/x²= c/x⁴
x²(x²– 2y²) = c (Where ‘c’ is integration constant)

Question 10. ye^x/ydx = (xe^x/y+ y)dy

Solution:

We have,
ye^x/ydx = (xe^x/y+ y)dy
(dy/dx) = (xe^x/y+ y)/ye^x/y
It is a homogeneous equation,
So, put x = vy (i)
On differentiating both sides w.r.t x,
dx/dy = v + y(dv/dy)
So,
v + y(dv/dy) = (vye^vy/y+ y)/ye^vy/y
v + y(dv/dy) = (ve^v+ 1)/e^v
y(dv/dy) = [(ve^v+ 1)/e^v] – v
y(dv/dy) = (ve^v+ 1 – ve^v)/e^v
y(dv/dy) = (1/e^v)
e^vdv = (dy/y)
On integrating both sides,
∫e^vdv = ∫(dy/y)
e^v= log(y) + log(c)
e^x/y= log(y) + log(c) (Where ‘c’ is integration constant)

Question 11. x²(dy/dx) = x²+ xy + y²

Solution:

We have,
x²(dy/dx) = x²+ xy + y²
dy/dx = (x²+ xy + y²)/x²
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) = (x²+ xvx + v²x²)/x²
v + x(dv/dx) = (1 + v + v²)
x(dv/dx) = (1 + v + v²) – v
dv/(1 + v²) = (dx/x)
On integrating both sides,
∫dv/(1 + v²) = ∫(dx/x)
tan^-1(v) = log|x| + c
tan^-1(y/x) = log|x| + c (Where ‘c’ is integration constant)

Question 12. (y²– 2xy)dx = (x²– 2xy)dy

Solution:

We have,
(y²– 2xy)dx = (x²– 2xy)dy
(dy/dx) = (y²– 2xy)/(x²– 2xy)
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– 2xvx)/(x²– 2xvx)
v + x(dv/dx) = (v²– 2v)/(1 – 2v)
x(dv/dx) = [(v²– 2v – v + 2v²)/(1 – 2v)]
x(dv/dx) = 3(v²– 1)/(1 – 2v)
-(2v – 1)dv/(v²– v) = 3(dx/x)
On integrating both sides,
-∫(2v – 1)dv/(v²– v) = 3∫(dx/x)
-log|v²– v| = 3log|x| – log|c|
log|v²– v| = log|c/x³|
(y²/x²– y/x) = (c/x³)
(y²– xy) = c/x
x(y²– xy) = c (Where ‘c’ is integration constant)

Question 13. 2xydx + (x²+ 2y²)dy = 0

Solution:

We have,

2xydx + (x²+ 2y²)dy = 0

dy/dx = -(2xy)/(x²+ 2y²)

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) = -(2xvx)/(x²+ 2v²x²)

v + x(dv/dx) = -(2v)/(1 + 2v²)

x(dv/dx) = -[(2v)/(1 + 2v²)] – v

$x\frac{dv}{dx}=\frac{-3v-2v^3}{1+2v^2}$

$\frac{1+2v^2}{3v+2v^3}dv=-\frac{dx}{x}$

On integrating both sides,

$∫\frac{1+2v^2}{3v+2v^3}dv=-∫\frac{dx}{x}$

Substituting (3v + 2v³) = z

On differentiating both sides w.r.t x,

3(1 + 2v)dv = dz

(1 + 2v)dv = (dz/3)

(1/3)∫(dz/z) = -∫(dx/x)

(1/3)log|z| = -log|x| + log|c|

log|3v + 2v³| = log|c/x|³

3y/x + 2(y/x)³= (c/x)³

(3yx²+ 2y³) = c (Where ‘c’ is integration constant)

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)

$∫\frac{2v+1}{2v(v+1)-1(v+1)}dx=-∫\frac{dx}{x}$

$∫\frac{2v+1}{(v+1)(2v-1)}dx=-∫\frac{dx}{x}$

Solving by partial fraction,

$\frac{2v+1}{(v+1)(2v-1)}=\frac{A}{(2v-1)}+\frac{B}{(v+1)}$

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)

$\frac{4}{3}∫\frac{dv}{(2v-1)}+\frac{1}{3}∫\frac{dv}{(v+1)}=-log|x|+log|c|$

(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)

$\frac{(2y-x)^2(x+y)}{x^2x}=\frac{c}{x^3}$

(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 + 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)|

$e^{2tan^{-1}(\frac{y}{x})}=\frac{(y^2+x^2)^{\frac{1}{2}}}{x}xc$

$e^{2tan^{-1}(\frac{y}{x})}={(y^2+x^2)^{\frac{1}{2}}}c$ (Where ‘c’ is integration constant)

Question 17. $\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}$

Solution:

We have,

$\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}$

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) = (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)

$\frac{y}{x}-[\sqrt{(\frac{y}{x})^2-1}]x=c$

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

$x\frac{dv}{dx}=\frac{-v^2-v+v^2-v^3}{1-v+v^2}$

$\frac{1-v+v^2}{-(v+v^3)}dv=\frac{dx}{x}$

$\frac{1-v+v^2}{-v(1+v^2)}dv=\frac{dx}{x}$

$(\frac{1}{1+v^2}-\frac{1}{v})dv=\frac{dx}{x}$

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|

$e^{tan^{-1}(\frac{y}{x})}=cy$ (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

$x\frac{dv}{dx}=\frac{v^2-\sqrt{1+v^2}-v^2}{v}$

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

$\frac{\frac{y}{x}cos\frac{y}{x}}{\frac{x}{y}sin\frac{y}{x}+cos\frac{y}{x}}$

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,

$v+x\frac{dv}{dx}=\frac{vcos(v)}{\frac{1}{v}sin(v)+cos(v)}$

$x\frac{dv}{dx}=\frac{v^2cos(v)}{sin(v)+vcos(v)}-v$

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

$\frac{dx}{dy}=\frac{x^2log(\frac{x}{y})-y^2}{xylog(\frac{x}{y})}$

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,

$v+y\frac{dv}{dy}=\frac{v^2y^2log(v)-y^2}{yvylog(v)}$

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

$\frac{dx}{dy}=-\frac{e^\frac{x}{y}(1-\frac{x}{y})}{1+e^\frac{x}{y}}$

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,

$v+y\frac{dv}{dy}=-\frac{e^v(1-v)}{1+e^v}$

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,

$∫\frac{1+v^2}{4(2-v)+v^2(2-v)}dv=∫\frac{dx}{x}$

$∫\frac{1+v^2}{(2-v)(4+v^2)}dv=∫\frac{dx}{x}$

Using partial fraction,

$∫\frac{1+v^2}{(2-v)(4+v^2)}=\frac{Av+B}{4+v^2}+\frac{C}{2-v}$

(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)

$∫\frac{-\frac{3}{8}v-\frac{3}{4}}{4+v^2}dv+∫\frac{\frac{5}{8}}{2-v}dv=∫\frac{dx}{x}$

$-\frac{3}{8}∫\frac{vdv}{v^2+1}-\frac{3}{4}∫\frac{dv}{v^2+1}+\frac{5}{8}∫\frac{dv}{2-v}=logx+logc$

$-\frac{3}{16}log|v^2+1|-\frac{3}{8}tan^{-1}(\frac{v}{2})-\frac{5}{8}log|2-v|=log|xc|$

$e^{-\frac{3}{8}tan^{-1}(\frac{v}{2})}=c|x(2-v)^\frac{5}{8}(v^2+4)^\frac{3}{16}|$

$e^{-\frac{3}{8}tan^{-1}(\frac{y}{2x})}=c|x(2-\frac{y}{x})^\frac{5}{8}(\frac{y^2}{x^2}+4)^\frac{3}{16}|$

$e^{-\frac{3}{8}tan^{-1}(\frac{y}{2x})}=c|x(2x-y)^\frac{5}{8}(y^2+4x^2)^\frac{3}{16}|$ (Where ‘c’ is integration constant)

Question 27. (x²– 2xy)dy + (x²– 3xy + 2y²)dx = 0

Solution:

We have,
(x²– 2xy)dy + (x²– 3xy + 2y²)dx = 0
(dy/dx) = (x²– 3xy + 2y²)/(2xy – 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) = (x²– 3xvx + 2v²x²)/(2xvx – x²)
v + x(dv/dx) = (1 – 3v + 2v²)/(2v – 1)
x(dv/dx) = [(1 – 3v + 2v²)/(2v – 1)] – v
x(dv/dx) = (1 – 3v + 2v²– 2v²+ v)/(2v – 1)
x(dv/dx) = (1 – 2v)/(2v – 1)
x(dv/dx) = -1
dv = -(dx/x)
On integrating both sides,
∫dv = -∫(dx/x)
v = -log|x| + log|c|
(y/x) + log|x| = log|c| (Where ‘c’ is integration constant)

Question 28. 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 29. x(dy/dx) – y = 2√(y²– x²)

Solution:

We have,

x(dy/dx) – y = 2√(y²– x²)

(dy/dx) = [2√(y²– x²) + 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,

$v+x\frac{dv}{dx}=\frac{2\sqrt{v^2x^2-x^2}+vx}{x}$

$v+x\frac{dv}{dx}=2\sqrt{v^2-1}+v$

x(dv/dx) = 2√(v²– 1)

dv/√(v²– 1) = 2(dx/x)

On integrating both sides,

∫dv/√(v²– 1) = 2∫(dx/x)

log|v + √(v²– 1)| = 2log(x) + log(c)

|v + √(v²– 1)| = |cx²|

$\frac{y}{x}+\sqrt{\frac{y^2}{x^2}-1}=|cx^2|$

$y+\sqrt{y^2-x^2}=cx^3$ (Where ‘c’ is integration constant)

Question 30. xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

Solution:

We have,

xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

xycos(y/x)dx + x²cos(y/x)dy = xysin(y/x)dy – y²sin(y/x)dx

x²cos(y/x)dy – xysin(y/x)dy = -y²sin(y/x)dx – xycos(y/x)dx

$\frac{dy}{dx}=\frac{xysin(\frac{y}{x})+y^2cos(\frac{y}{x})}{xysin(\frac{y}{x})-x^2cos(\frac{y}{x})}$

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,

$v+x\frac{dv}{dx}=\frac{xvxsinv+v^2x^2cosv}{xvxsinv-x^2cosv}$

v + x(dv/dx) = (vcosv + v²sinv)/(vsinv – cosv)

x(dv/dx) = [(vcosv + v²sinv)/(vsinv – cosv)] – v

x(dv/dx) = (vcosv + v²sinv – v²sinv + vcosv)/(vsinv – cosv)

x(dv/dx) = (2vcosv)/(vsinv – cosv)

[(vsinv – cosv)/(vcosv)]dv = 2(dx/x)

On integrating both sides,

∫tanvdv – ∫(dv/v) = 2log|x| + log|c|

log|secv| – log|v| = log|cx²|

log|(secv/v)| = log|cx²|

(x/y)sec(y/x) = cx²

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

Question 31. (x²+ 3xy + y²)dx – x²dy = 0

Solution:

We have,

(x²+ 3xy + y²)dx – x²dy = 0

dy/dx = (x²+ 3xy + y²)/x²

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) = (x²+ 3xvx + v²x²)/x²

v + x(dv/dx) = (1 + 3v + v²)

x(dv/dx) = (1 + 3v + v²) – v

x(dv/dx) = (1 + 2v + v²)

x(dv/dx) = (1 + v)²

dv/(1 + v)²= (dx/x)

On integrating both sides,

∫dv/(1 + v)²= ∫(dx/x)

-[1/(v + 1)] = log|x| – c

$-\frac{1}{\frac{y}{x}+1}=log|x|-c$

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

Question 32. (x – y)(dy/dx) = (x + 2y)

Solution:

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(x – 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)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v²)/(1 – v)

x(dv/dx) = (1 + v + v²)/(1 – v)

(1 – v)dv/(1 + v + v²) = (dx/x)

On integrating both sides,

∫[(1 – v)/(1 + v + v²)]dv = ∫(dx/x)

$-\frac{1}{2}∫\frac{2v-2}{v^2+v+1}dv=∫\frac{dx}{x}$

$∫\frac{(2v+1)-3}{v^2+v+1}dv=-2∫\frac{dx}{x}$

$∫\frac{2v+1}{v^2+v+1}dv-∫\frac{3dv}{v^2+2v(\frac{1}{2})+(\frac{1}{2})^2-(\frac{1}{2})^2+1}=-2log|x|+c$

$log|v^2+v+1|-∫\frac{3dv}{(v+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=-log|x^2|+c$

$log|\frac{y^2}{x^2}+\frac{y}{x}+1|-3(\frac{2}{\sqrt{3}})tan^{-1}(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}})+log|x^2|=c$

$log|y^2+xy+x^2|=2\sqrt{3}tan^{-1}(\frac{2y+x}{x\sqrt{3}})+c$ (Where ‘c’ is an integration constant)

Question 33. (2x²y + y³)dx + (xy²– 3x²)dy = 0

Solution:

We have,

(2x²y + y³)dx + (xy²– 3x²)dy = 0

dy/dx = (2x²y + y³)/(3x³– xy²)

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) = (2x²vx + v³x³)/(3x³– xv²x²)

v + x(dv/dx) = (2v + v³)/(3 – v³)

x(dv/dx) = [(2v + v³)/(3 – v³)] – v

x(dv/dx) = (2v + v³– 3v + v³)/(3 – v³)

(3 – v³)dv/(2v³– v) = (dx/x)

On integrating both sides,

∫[(3 – v³)/(2v³– v)]dv = ∫(dx/x)

$∫\frac{3-v^2}{v(2v^2-1)}dv=∫\frac{dx}{x}$

Using partial fraction,

$\frac{3-v^2}{v(2v^2-1)}=\frac{A}{v}+\frac{Bv+C}{2v^2-1}$

3 – v²= A(2v²– 1) + (Bv + C)v

3 – v²= 2Av²– A + Bv²+ Cv

3 – v²= v²(2A + B) + Cv – A

On comparing the coefficients, we get

A = -3,

B = 5,

C = 0,

$-3∫\frac{dv}{v}+∫\frac{5v}{2v^2-1}dv=∫\frac{dx}{x}$

$-3∫\frac{dv}{v}+\frac{5}{4}∫\frac{4v}{2v^2-1}dv=∫\frac{dx}{x}$

-3log|v|+(5/4)log|2v²-1|=log|x|+log|c|

-12log|v|+5log|2v²-1|=4log|x|+4log|c|

$log|\frac{2v^2-1}{v^{12}}|=log|x^4c^4|$

$(2\frac{y^2}{x^2}-1)^5=x^4c^4\frac{y^{12}}{x^{12}}$

$\frac{2y^2-x^2}{x^{10}}=x^4c^4\frac{y^{12}}{x^{12}}$

|2y²– x²|⁵= x²c⁴y¹² (Where ‘c’ is an integration constant)

Question 34. x(dy/dx) – y + xsin(y/x) = 0

Solution:

We have,
x(dy/dx) – y + xsin(y/x) = 0
x(dy/dx) = y – xsin(y/x)
(dy/dx) = [y – xsin(y/x)]/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) = [vx – xsinv]/x
v + x(dv/dx) = (v – sinv)
x(dv/dx) = -sinv
cosecvdv = -(dx/x)
On integrating both sides,
∫cosecvdv = -∫(dx/x)
-log|cosecv + cotv| = -log|x| + log|c|
-log|(1/sinv) + (cosv/sinv)| = -log|x/c|
|(1 + cosv)/sinv| = |x/c|
xsinv = c(1 + cosv)
xsin(y/x) = c[1 + cos(y/x)] (Where ‘c’ is integration constant)

Question 35. ydx + {xlog(y/x)}dy – 2xydy = 0

Solution:

We have,

ydx + {xlog(y/x)}dy – 2xydy = 0

y + {xlog(y/x)}(dy/dx) – 2xy = 0

$\frac{dy}{dx}=\frac{y}{2x-xlog(\frac{y}{x})}$

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,

$v+x\frac{dv}{dx}=-\frac{vx}{2x-xlogv}$

v + x(dv/dx) = v/(2 – logv)

x(dv/dx) = [v/(2 – logv)] – v

x(dv/dx) = (v – 2v + vlogv)/(2 – logv)

x(dv/dx) = -v(logv – 1)/(logv – 2)

$\frac{(logv-2)}{v(logv-1)}dv=-\frac{dx}{x}$

On integrating both sides,

$∫\frac{(logv-2)}{v(logv-1)}dv=-∫\frac{dx}{x}$

$∫\frac{(logv-1)-1}{v(logv-1)}dv=-∫\frac{dx}{x}$

$∫\frac{dv}{v(logv-1)}-∫\frac{dv}{v(logv-1)}=-∫\frac{dx}{x}$

Let. logv – 1 = z

On differentiating both sides,

(dv/v) = dz

∫dz – ∫(dz/z) = -∫(dx/x)

z – log|z| = -log|x| + log|c|

(logv – 1) – log|(logv – 1)| = -log|x| + log|c|

logv – log|logv – 1| = -log|x| + log|c| + 1

log|(logv – 1)/v| = log|c₁x|

|logv – 1| = |c₁xv|

|log(y/x) – 1| = |c₁x(y/x)|

|log(y/x) – 1| = |c₁y| (Where ‘c₁’ is integration constant)

Question 36(i). (x²+ y²)dx = 2xydy, y(1) = 0

Solution:

We have,
(x²+ y²)dx = 2xydy
(dy/dx) = (x²+ y²)/2xy
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) = (x²+ v²x²)/2vx²
v + x(dv/dx) = (1 + v²)/2v
x(dv/dx) = [(1 + v²)/2v] – v
x(dv/dx) = (1 + v²– 2v²)/2v
x(dv/dx) = (1 – v²)/2v
2vdv/(1 – v²) = (dx/x)
On integrating both sides,
∫2vdv/(1 – v²) = ∫(dx/x)
-log|1 – v²| = log|x| – log|c|
log|1 – v²| = log|c/x|
|1 – y²/x²| = |c/x|
|x²– y²| = |cx|
At x = 1, y = 0
1 – 0 = c
c = 1
|x²– y²| = |x|
(x²– y²)= x

Question 36(ii). xe^x/y– y + x(dy/dx) = 0, y(e) = 0

Solution:

We have,
xe^x/y– y + x(dy/dx) = 0
(dy/dx) = (y – xe^x/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) = (vx – xe^v)/x
v + x(dv/dx) = v – e^v
x(dv/dx) = v – e^v– v
x(dv/dx) = -e^v
e^-vdv = -(dx/x)
On integrating both sides,
∫e^-vdv = -∫(dx/x)
-e^-v= -log|x| – log|c|
e^-v= log|x| + log|c|
e^-(y/x)= log|x| + log|c|
At x = e, y = 0
e^-(0/e)= log|e| + log|c|
1 = 1 + log|c|
c = 0
e^-y/x= logx

Question 36(iii). (dy/dx) – (y/x) + cosec(y/x) = 0, y(1) = 0

Solution:

We have,
(dy/dx) – (y/x) + cosec(y/x) = 0
(dy/dx) = (y/x) – cosec(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 – cosec(v)
x(dv/dx) = v – cosec(v) – v
x(dv/dx) = -cosec(v)
-sin(v)dv = (dx/x)
On integrating both sides,
-∫sin(v)dv = ∫(dx/x)
cos(v) = log|x| + log|c|
cos(y/x) = log|x| + log|c|
At x = 1, y = 0
cos(0/1) = log|1| + log|c|
1 = 0 + log|c|
log|c| = 1
cos(y/x) = log|x| + 1
log|x| = cos(y/x) – 1

Question 36(iv). (xy – y²)dx – x²dy = 0, y(1) = 1

Solution:

We have,
(xy – y²)dx – x²dy = 0
(dy/dx) = (xy – y²)/x²
(dy/dx) = (y/x) – (y²/x²)
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 – v²
x(dv/dx) = v – v²– v
x(dv/dx) = -v²
-(dv/v²) = (dx/x)
On integrating both sides,
-∫(dv/v²) = ∫(dx/x)
-(-1/v) = log|x| + c
(1/v) = log|x| + c
x/y = log|x| + c
At x = 1, y = 1
1 = log|1| + c
c = 1
x/y = log|x| + 1
y = x/[log|x| + 1]

Question 36(v). (dy/dx) = [y(x + 2y)]/[x(2x + y)]

Solution:

We have,

(dy/dx) = [y(x + 2y)]/[x(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) = [vx(x + 2vx)]/[x(2x + vx]

x(dv/dx) = [vx(x + 2vx)]/[x(2x + vx] – v

x(dv/dx) = (v + 2v²– 2v – v²)/(2 + v)

x(dv/dx) = (v²– v)/(2 + v)

(2 + v)dv/[v(v – 1)] = (dx/x)

On integrating both sides,

$∫\frac{2+v}{v(v-1)}dv=\frac{dx}{x}$

Using partial derivative,

$\frac{2+v}{v(v-1)}dv=\frac{A}{v}+\frac{B}{v-1}$

2 + v = A(v – 1) + B(v)

2 + v = v(A + B) – A

On comparing the coefficients,

A = -2

B = 3

-2∫(dv/v) + 3∫dv/(v – 1) = ∫(dx/x)

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

log|(v – 1)³/v²| = log|xc|

(v – 1)³= v²|xc|

(y – x)³/x³= (y/x)²|xc|

(y – x)³ = y²x²c

At x = 1, y = 2,

(2 – 1)³ = 4 * 1 * c

c = (1/4)

(y – x)³ = (1/4)y²x²

Question 36(vi). (y⁴– 2x³y)dx + (x⁴– 2xy³)dy = 0, y(1) = 0

Solution:

We have,

(y⁴– 2x³y)dx + (x⁴– 2xy³)dy = 0

dy/dx = (2x³y – y⁴)/(x⁴– 2xy³)

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) = (2x³vx – v⁴x⁴)/(x⁴– 2xv³x³)

v + x(dv/dx) = (2v – v⁴)/(1 – 2v³)

x(dv/dx) = [(2v – v⁴)/(1 – 2v³)] – v

x(dv/dx) = (2v – v⁴– v + 2v⁴)/(1 – 2v³)

x(dv/dx) = (v + v⁴)/(1 – 2v³)

$\frac{1-2v^3}{v(1+v^3)}dv=\frac{dx}{x}$

$\frac{1+v^3-3v^3}{v(1+v^3)}dv=\frac{dx}{x}$

On integrating both sides,

$∫\frac{1+v^3}{v(1+v^3)}dv-∫\frac{3v^3}{v(1+v^3)}dv=∫\frac{dx}{x}$

∫(dv/v) – ∫(3v²)dv/(1 + v³) = log|x| + log|c|

log|v| – log|1 + v³| = log|xc|

log|v/(1 + v³)| = log|xc|

$\frac{\frac{y}{x}}{1+(\frac{y}{x})^3}=|cx|$

At x = 1, y = 1,

1/(1 + 1) = c

c = (1/2)

(yx²)/(x³+ y³) = (1/2)x

Question 36(vii). x(x²+ 3y²)dx + y(y²+ 3x²)dy = 0, y(1) = 1

Solution:

We have,

x(x²+ 3y²)dx + y(y²+ 3x²)dy = 0

dy/dx = -[x(x²+ 3y²)/y(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,

$v+x\frac{dv}{dx}=-\frac{x(x^2+3v^2x^2)}{vx(v^2x^2+3x^2)}$

$v+x\frac{dv}{dx}=-\frac{(1+3v^2)}{v(v^2+3)}$

$x\frac{dv}{dx}=-\frac{(1+3v^2)}{v(v^2+3)}-v$

x(dv/dx) = -(1 + 3v²+ v⁴+ 3v²)/v(v²+ 3)

[(v³+ 3v)/(1 + 6v²+ v⁴)]dv = -(dx/x)

Multiply both sides with 4 and integrating,

$∫\frac{4v^3+12v}{v^4+6v^2+1}dv=-4∫\frac{dx}{x}$

log|v⁴+ 6v²+ 1| = -log|x|⁴+ log|c|

|v⁴+ 6v²+ 1| = |c/x⁴|

(y⁴+ 6x²y²+ x⁴) = c

At y = 1, x = 1

(1 + 6 + 1) = c

c = 8

(y⁴+ 6x²y²+ x⁴) = 8

Question 36(viii). {xsin²(y/x) – y}dx + xdy = 0, y(1) = π/4

Solution:

We have,
{xsin²(y/x) – y}dx + xdy = 0
dy/dx = [y – xsin²(y/x)]/x
dy/dx = (y/x) – sin²(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 – sin²(v)
x(dv/dx) = v – sin²(v) – v
x(dv/dx) = -sin²(v)
-cosec²(v)dv = (dx/x)
On integrating both sides,
-∫cosec²(v) = ∫(dx/x)
cot(v) = log|x| + log|c|
cot(y/x) = log|xc|
At x = 1, y = π/4
cot(π/4) = log|c|
log|c| = 1
c = e
cot(y/x) = log|ex|

Question 36(ix). x(dy/dx) – y + xsin(y/x) = 0, y(2) = π

Solution:

We have,
x(dy/dx) – y + xsin(y/x) = 0
x(dy/dx) = y – xsin(y/x)
(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) = v – sin(v) – v
x(dv/dx) = -sin(v)
-cosec(v)dv = (dx/x)
On integrating both sides,
∫cosec(v) = -∫(dx/x)
log|cosec(v) – cot(v)| = -log|x| + log|c|
log|cosec(v) – cot(v)| = -log|x| + log|c|
log|cosec(y/x) – cot(y/x)| = -log|x| + log|c|
At x = 2, y = π
|cosec(π/2) – cot(π/2)| = -log|2| + log|c|
log|c| = 0
log|cosec(y/x) – cot(y/x)| = -log|x|

Question 37. xcos(y/x)(dy/dx) = ycos(y/x) + x, When x = 1, y = π/4

Solution:

We have,

xcos(y/x)(dy/dx) = ycos(y/x) + x

$\frac{dy}{dx}=\frac{ycos(\frac{y}{x})+x}{xcos(\frac{y}{x})}$

(dy/dx) = (y/x) + [1/cos(y/x)]

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 + 1/cosv

x(dv/dx) = v + 1/cosv – v

x(dv/dx) = 1/cosv

cosvdv = (dx/x)

On integrating both sides,

∫cosvdv = ∫(dx/x)

sin(v) = log|x| + log|c|

sin(y/x) = log|x| + log|c|

At x = 1, y = π/4

1/√2 = 0 + log|c|

log|c| = (1/√2)

sin(y/x) = log|x| + (1/√2)

Question 38. (x – y)(dy/dx) = (x + 2y), when x = 1,y = 0

Solution:

We have,

(x – y)(dy/dx) = (x + 2y)

(dy/dx) = (x + 2y)/(x – 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)/(x – vx)

v + x(dv/dx) = (1 + 2v)/(1 – v)

x(dv/dx) = [(1 + 2v)/(1 – v)] – v

x(dv/dx) = (1 + 2v – v + v²)/(1 – v)

(1 – v)dv/(1 + v + v²) = (dx/x)

On integrating both sides,

$∫\frac{1-v}{1+v+v^2}dv=\frac{dx}{x}$

$\frac{1}{2}∫\frac{2-2v}{1+v+v^2}dv=∫\frac{dx}{x}$

$∫\frac{3-(1+2v)}{1+v+v^2}dv=∫\frac{dx}{x}$

$\frac{3}{2}∫\frac{dv}{1+v+v^2}-\frac{1}{2}∫\frac{2v+1}{1+v+v^2}dv=∫\frac{dx}{x}$

$\frac{3}{2}∫\frac{dv}{(v+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}-\frac{1}{2}log|v^2+v+1|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}})-\frac{1}{2}log|v^2+v+1|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|\frac{x^2+xy+y^2}{x^2}|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|x^2+xy+y^2|+log|x|=log|x|+c$

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|x^2+xy+y^2|=c$

At x = 1, y = 0

√3tan^-1|1/√3| – (1/2)log|1| = c

c = √3(π/6)

c = (π/2√3)

$\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{1}{2}log|x^2+xy+y^2|=\frac{π}{2\sqrt{3}}$

$\frac{1}{2}log|x^2+xy+y^2|=2\sqrt{3}tan^{-1}(\frac{2y+1}{\sqrt{3}x})-\frac{π}{\sqrt{3}}$

Question 39. (dy/dx) = xy/(x²+ y²)

Solution:

We have,
(dy/dx) = xy/(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) = xvx/(x²+ v²x²)
v + x(dv/dx) = v/(1 + v²)
x(dv/dx) = [v/(1 + v²)] – v
x(dv/dx) = (v – v – v³)/(1 + v²)
[-(1/v³) – (1/v)]dv = (dx/x)
On integrating both sides,
-∫dv/v³– ∫dv/v = ∫(dx/x)
(1/2v²) – log|v| = log|x| + c
(x²/2y²) = log|vx| + c
(x²/2y²) = log|(y/x)x| + c
(x²/2y²) = log|y| + c
At x = 0, y = 1
c = 0
(x²/2y²) = log|y|
Solve the following differential equations:
Question 1. dy/dx + 2y = e^3x
Solution:

We have,
dy/dx + 2y = e^3x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = e^3x
So, I.F = e^∫Pdx
= e^∫2dx
= e^2x
The solution of differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = ∫e^3x.e^2xdx + c
y(e^2x) = (1/5)e^5x+ c
y = (e^3x/5) + ce^-2x
Hence, this is the required solution.
Question 2. 4(dy/dx) + 8y = 5e^-3x
Solution:
We have,
4(dy/dx) + 8y = 5e^-3x
(dy/dx) + 2y = (5/4)e^-3x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = (5/4)e^-3x
So, I.F = e^∫Pdx
= e^∫2dx
= e^2x
The solution of differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = (5/4)∫e^-3x.e^2xdx + c
y(e^2x) = (5/4)∫e^-xdx + c
y = -(5/4)e^-3x+ ce^-2x
This is the required solution.
Question 3. (dy/dx) + 2y = 6e^x
Solution:
We have,
(dy/dx) + 2y = 6e^x ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = 6e^x
So, I.F = e^∫Pdx
= e^∫2dx
= e^2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = ∫6e^x.e^2xdx + c
y(e^2x) = 6∫e^3xdx + c
y(e^2x) = 2e^3x+ c
y = 2e^x+ ce^-2x
This is the required solution.
Question 4. (dy/dx) + y = e^-2x
Solution:
We have,
(dy/dx) + y = e^-2x ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1, Q = e^-2x
So, I.F = e^∫Pdx
= e^∫dx
= e^x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^x) = ∫e^-2x.e^xdx + c
y(e^x) = ∫e^-xdx + c
y(e^x) = -e^-x+ c
y = -e^-2x+ ce^-x
This is the required solution.
Question 5. x(dy/dx) = x + y
Solution:
We have,
x(dy/dx) = x + y
(dy/dx) = 1 + (y/x)
(dy/dx) – (y/x) = 1 ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = (-1/x), Q = 1
So, I.F = e^∫Pdx
= e^-∫(dx/x)
= e^-log(x)
= e^log(1/x)
= (1/x)
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/x) = ∫(1/x)dx + c
(y/x) = log|x| + c
y = xlog|x| + cx
This is the required solution.
Question 6. (dy/dx) + 2y = 4x
Solution:
We have,
(dy/dx) + 2y = 4x ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = 4x
So,
I.F = e^∫Pdx
= e^∫2dx
= e^2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = ∫4x.e^2xdx + c
y(e^2x) = 4x∫e^2xdx – 4∫{(dx/dx)∫e^2xdx}dx + c
y(e^2x) = 2xe^2x– 2∫e^2xdx + c
y(e^2x) = 2xe^2x– e^2x+ c
y = (2x – 1) + ce^-2x
This is the required solution.
Question 7. x(dy/dx) + y = xe^x
Solution:
We have,
x(dy/dx) + y = xe^x
(dy/dx) + (y/x) = e^x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = (1/x), Q = e^x
So,
I.F = e^∫Pdx
= e^∫(dx/x)
= e^log(x)
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x) = ∫x.e^xdx + c
xy = x∫e^xdx – {(dx/dx)∫e^xdx}dx + c
xy = xe^x– ∫e^xdx + c
xy = xe^x– e^x+ c
$y=\frac{x-1}{x}e^x+\frac{c}{x}$
This is the required solution.
Question 8. (dy/dx) + [4x/(x²+ 1)]y = -1/(x²+ 1)²
Solution:
We have,
(dy/dx) + [4x/(x²+ 1)]y = -1/(x²+ 1)²   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = [4x/(x²+ 1)], Q = -1/(x²+ 1)²
So,
I.F = e^∫Pdx
$=e^{∫\frac{4x}{x^2+1}dx}$
$=e^{2log|x^2+1|}$
= (x²+1)²
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x²+ 1)²= ∫-[1/(x²+ 1)²](x²+ 1)²dx + c
y(x²+ 1)²= -∫dx + c
y(x²+ 1)²= -x + c
y = -x/(x²+ 1)²+ c/(x²+ 1)²
This is the required solution.
Question 9. x(dy/dx) + y = xlogx
Solution:
We have,
x(dy/dx) + y = xlogx
(dy/dx) + (y/x) = logx ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = (1/x), Q = logx
So,
I.F = e^∫Pdx
= e^∫(dx/x)
= e^log(x)
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x) = ∫x.logxdx + c
xy = logx∫xdx – {(d/dx)logx∫xdx}dx + c
xy = (x²/2)logx – ∫(1/x)(x²/2) + c
xy = (x²/2)logx – (1/2)∫xdx + c
xy = (x²/2)logx – (x²/4) + c
y = (x/2)logx – (x/4) + (c/x)
This is the required solution.
Question 10. x(dy/dx) – y = (x – 1)e^x
Solution:
We have,
x(dy/dx) – y = (x – 1)e^x
(dy/dx) – (y/x) = [(x – 1)/x]e^x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -(1/x), Q = [(x – 1)/x]e^x
So,
I.F = e^∫Pdx
= e^-∫(dx/x)
= e^-log(x)
= e^log(1/x)
= 1/x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
$y(\frac{1}{x})=∫(\frac{x-1}{x})e^x.(\frac{dx}{x})+c$
(y/x) = ∫[(1/x) – (1/x²)]e^x+ c
Since, ∫[f(x) + f'(x)]e^xdx = f(x)e^x+ c
(y/x) = (e^x/x) + c
y = e^x+ xc
This is the required solution.
Question 11. (dy/dx) + y/x = x³
Solution:
We have,
(dy/dx) + y/x = x³   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/x, Q = x³
So, I.F = e^∫Pdx
= e^∫dx/x
= e^logx
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
yx = ∫x³.xdx + c
yx = ∫x⁴dx + c
yx = (x⁵/5) + c
y = (x⁴/5) + c/x
This is the required solution.
Question 12. (dy/dx) + y = sinx
Solution:
We have,
(dy/dx) + y = sinx ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1, Q = sinx
So, I.F = e^∫Pdx
= e^∫dx
= e^x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^x) = ∫sinx.e^xdx + c
Let, I = ∫sinx.e^xdx
I = e^x∫sinxdx – ∫{(d/dx)e^x∫sinxdx}dx
I = -e^xcos + ∫e^xcosxdx
I = -e^xcosx + e^x∫cosxdx – ∫{(d/dx)e^x∫cosxdx}dx
I = -e^xcosx + e^xsinx – ∫e^xsinxdx
2I = e^x(sinx – cosx)
I = (e^x/2)(sinx – cosx)
y(e^x) = (e^x/2)(sinx – cosx) + c
y = (1/2)(sinx – cosx) + ce^-x
This is the required solution.
Question 13. (dy/dx) + y = cosx
Solution:
We have,
(dy/dx) + y = cosx ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1, Q = cosx
So, I.F = e^∫Pdx
= e^∫dx
= e^x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^x) = ∫cosx.e^xdx + c
Let, I = ∫cosx.e^xdx
I = e^x∫cosxdx – ∫{(d/dx)e^x∫cosxdx}dx
I = e^xsinx – ∫e^xsinxdx
I = e^xsinx – e^x∫sinxdx + ∫{(d/dx)e^x∫sinxdx}dx
I = e^xsinx + e^xcosx – ∫e^xcosxdx
2I = e^x(cosx + sinx)
I = (e^x/2)(cosx + sinx)
y(e^x) = (e^x/2)(cosx + sinx) + c
y = (1/2)(cosx + sinx) + ce^-x
This is the required solution.
Question 14. (dy/dx) + 2y = sinx
Solution:
We have,
(dy/dx) + 2y = sinx ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = sinx
So, I.F = e^∫Pdx
= e^∫2dx
= e^2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = ∫sinx.e^2xdx + c
Let, I = ∫sinx.e^2xdx
I = e^2x∫sinxdx – {(d/dx)e^2x∫sinxdx}dx
I = -e^2xcosx + 2∫e^2xcosdx
I = -e^2xcosx + 2e^2x∫cosxdx – 2{(d/dx)e^2x∫cosxdx}dx
I = -e^2xcosx + 2e^2xsinx – 4∫e^2xsinxdx
5I = e^2x(2sinx – cosx)
I = (e^2x/5)(2sinx – cosx)
y(e^2x) = (e^2x/5)(2sinx – cosx) + c
y = (1/5)(2sinx – cosx) + ce^-2x
This is the required solution.
Question 15. (dy/dx) – ytanx = -2sinx
Solution:
We have,
(dy/dx) – ytanx = -2sinx ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -tanx, Q = sinx
So,
I.F = e^∫Pdx
= e^∫-tanxdx
= e^-log|secx|
= 1/secx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/secx) = -2∫sinx.(1/secx)dx + c
ycosx = -∫2sinx.cosxdx + c
ycosx = -∫sin2xdx + c
ycosx = (cos2x/2) + c
y = (cos2x/cosx) + (c/cosx)
This is the required solution.
Question 16. (1 + x²)(dy/dx) + y = tan^-1x
Solution:
We have,
(1 + x²)(dy/dx) + y = tan^-1x
(dy/dx) + [y/(1 + x²)] = tan^-1x/(1 + x²) ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/(1 + x²), Q = tan^-1x/(1 + x²)
So,
I.F = e^∫Pdx
$=e^{∫\frac{dx}{1+x^2}}$
$=e^{tan^{-1}x}$
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
$y(e^{tan^{-1}x})=∫\frac{tan^{-1}x}{1+x^2}e^{tan^{-1}x}dx+c$
Let, tan^-1x = z
On differentiating both sides we get,
dx/(1 + x²) = dz
y(e^z) = ∫ze^zdz + c
y(e^z) = z∫e^zdz – {(dz/dz)∫e^zdz}dz
y(e^z) = ze^z– ∫ e^zdz + c
y(e^z) = e^z(z – 1) + c
y = (z – 1) + ce^-z
$y=(tan^{-1}x-1)+ce^{tan^{-1}x}$
This is the required solution.
Question 17. (dy/dx) + ytanx = cosx
Solution:
We have,
(dy/dx) + ytanx = cosx ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = tanx, Q = cosx
So, I.F = e^∫Pdx
= e^∫tanxdx
= e^log|secx|
= secx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y.secx = ∫cosx.secxdx + c
y.secx = ∫dx + c
y.secx = x + c
y = xcosx + c.cosx
This is the required solution.
Question 18. (dy/dx) + ycotx = x²cotx + 2x
Solution:
We have,
(dy/dx) + ycotx = x²cotx + 2x ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cotx, Q = x²cotx + 2x
So,
I.F = e^∫Pdx
= e^∫cotxdx
= e^log|sinx|
= sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(sinx) = ∫(x²cotx + 2x)sinxdx + c
y(sinx) = ∫(x²cosx + 2xsinx)dx + c
y(sinx) = x²∫cosxdx – {(d/dx)x²∫cosxdx}dx + ∫2xsinxdx + c
y(sinx) = x²sinx – ∫2xsinxdx + ∫2xsinxdx + c
y(sinx) = x²sinxdx + c
This is the required solution.
Question 19. (dy/dx) + ytanx = x²cos²x
Solution:
We have,
(dy/dx) + ytanx = x²cos²x ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = tanx, Q = x²cos²x
So,
I.F = e^∫Pdx
= e^∫tanxdx
= e^log|secx|
= secx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(secx) = ∫secx.(x².cos²x)dx + c
y(secx) = ∫x².cosxdx + c
y(secx) = x²∫cosxdx – ∫{(d/dx)x²∫cosxdx}dx + c
y(secx) = x²sinx – 2∫xsinxdx + c
y(secx) = x²sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c
y(secx) = x²sinx + 2xcosx – 2∫cosxdx + c
y(secx) = x²sinx + 2xcosx – 2sinx + c
This is the required solution.
Question 20. (1 + x²)(dy/dx) + y = $e^{tan^{-1}x}$
Solution:
We have,
(1 + x²)(dy/dx) + y = $e^{tan^{-1}x}$
(dy/dx) + [1/(x²+ 1)]y = $e^{tan^{-1}x}$ /(x²+ 1) ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/(x²+ 1), Q = $e^{tan^{-1}x}$ /(x² + 1)
So,
I.F = e^∫Pdx
$=e^{∫\frac{1}{x^2+1}dx}$
= $e^{tan^{-1}x}$
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y( $e^{tan^{-1}x}$ ) – ∫[ $e^{tan^{-1}x}$ /(x²+ 1)] $e^{tan^{-1}x}$ dx + c
Let, tan^-1x = z
On differentiating both sides we get
dx/(1 + x²) = dz
ye^z= ∫e^2zdz + c
ye^z= (e^2z/2) + c
y = (e^z/z) + c.e^-z
y = (1/2) $e^{tan^{-1}x}$ + c. $e^{tan^{-1}x}$
This is the required solution.
Question 21. xdy = (2y + 2x⁴+ x²)dx
Solution:
We have,
xdy = (2y + 2x⁴+ x²)dx
(dy/dx) = 2(y/x) + 2x³+ x
(dy/dx) – (y/x) = 2x³+ x ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -(2/x), Q = 2x³+ x
So,
I.F = e^∫Pdx
= e^-2∫(dx/x)
= e^-2log(x)
= e^2log|1/x|
= 1/x²
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/x²) = ∫(1/x²).(2x³+ x)dx + c
(y/x²) = ∫[2x + (1/x)]dx + c
(y/x²) = x²+ log|x| + c
y = x⁴+ x²log|x| + cx²
This is the required solution.
Question 22. (1 + y²) + (x – $e^{tan^{-1}y}$ )(dy/dx) = 0
Solution:
We have,
(1 + y²) + (x – $e^{tan^{-1}y}$ )(dy/dx) = 0
$\frac{dy}{dx}=-\frac{1+y^2}{x-e^{tan^{-1}}x}$
$\frac{dx}{dy}=-\frac{x-e^{tan^{-1}}x}{1+y^2}$
(dx/dy) + x/(1 + y²) = $e^{tan^{-1}y}$ /(1 + y²) ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = 1/(1 + y²), Q = $e^{tan^{-1}y}$ /(1 + y²)
So,
I.F = e^∫Pdy
$e^{∫\frac{dy}{1+y^2}}$
= $e^{tan^{-1}y}$
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
$x(e^{tan^{-1}y})=∫(e^{tan^{-1}y})(\frac{e^{tan^{-1}y}}{1+y^2})dy+c$
Let, tan^-1y = z
On differentiating both sides we have,
dy/(1 + y²) = dz
xe^z= ∫e^2zdz + c
xez = (e2z/2) + c
x = e^z/2 + ce^-z
$x=\frac{e^{tan^{-1}y}}{2}+c.e^{tan^{-1}y}$
This is the required solution.
Question 23. y²(dx/dy) + x – 1/y = 0
Solution:
We have,
y²(dx/dy) + x – 1/y = 0
(dx/dy) + (x/y²) = 1/y³   ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = 1/y², Q = 1/y³
So,
I.F = e^∫Pdy
= e^∫dy/y2
= e^-(1/y)
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
e^-(1/y)x = ∫(1/y³).(e^-1/y)dy + c
Let,-(1/y) = z
Differentiating both sides we have,
(dy/y²) = dz
xe^z= -∫ze^zdz + c
xe^z= -z∫e^zdz + ∫{(dz/dz)∫e^zdz}dz + c
xe^z= -ze^z+ ∫e^zdz + c
xe^z= -ze^z+ e^z+ c
x = (1 – z) + ce – z
x = [1 + (1/y)] + ce^1/y
x = (y + 1)/y + ce^1/y
This is the required solution.

Solve the following differential equations:
Question 24. (2x – 10y³)(dy/dx) + y = 0
Solution:

We have,
(2x – 10y³)(dy/dx) + y = 0
(2x – 10y³)(dy/dx) = -y
(dx/dy) = -(2x – 10y³)/y
(dx/dy) + 2x/y = 10y²  ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = 2/y, Q = 10y²
So,
I.F = e^∫Pdy
= e^∫(2/y)dy
= e^2log|y|
= y²
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(y²) = ∫(10y²)(y²)dy + c
xy² = 10(y⁵/5) + c
x = 2y³+ cy^-2
This is the required solution.
Question 25. (x + tany)dy = sin2ydx
Solution:
We have,
(x + tany)dy = sin2ydx
(dx/dy) = (x + tany)/sin2y
(dx/dy) – cosec2y.x = tany/sin2y ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = -cosec2y, Q = tany/sin2y
So, I.F = e^∫Pdy
= e^{∫-cosec2ydy}
= $e^{-\frac{1}{2}log|tany|}$
= $e^{log|\sqrt{coty}|}$
= √coty
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(√coty) = ∫(tany/sin2y).(√coty)dy + c
$x\sqrt{coty}=∫\frac{\sqrt{tany}}{\frac{2tany}{1+tan^2y}}dy+c$
$x\frac{1}{\sqrt{tany}}=∫\frac{1+tan^2y}{2\sqrt{tany}}dy+c$
$x\frac{1}{\sqrt{tany}}=∫\frac{sec^2x}{2\sqrt{tany}}dy+c$
Let, tany = z
On differentiating both side we have,
sec²ydx = dz
(x/√tany) = (1/2)∫dz/√z + c
(x/√tany) = (1/2)(2√z) + c
x = (√tany)(√tany) + c(√tany)
x = tany + c(√tany)
This is the required solution.
Question 26. dx + xdy = e^-ysec²ydy
Solution:
We have,
dx + xdy = e^-ysec²ydy
(x – e^-ysec²y)dy = -dx
(dx/dy) = (e^-ysec²y-x) ………..(i)
The above equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = 1, Q = e^-ysec²y
So, I.F = e^∫Pdy
= e^∫dy
= e^y
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(e^y) = ∫e^-ysce²ye^ydy + c
x(e^y) = ∫sec²ydy + c
x(e^y) = tany + c
x = (tany + c)e^-y
This is the required solution.
Question 27. (dy/dx) = ytanx – 2sinx
Solution:
We have,
(dy/dx) = ytanx – 2sinx
(dy/dx) – ytanx = -2sinx   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -tanx, Q = sinx
So,
I.F = e^∫Pdx
= e^∫-tanxdx
= e^-log|secx|
= 1/secx
= cosx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(cosx) = -2∫sinx.(cosx)dx + c
ycosx = -∫2sinx.cosxdx + c
Let, sinx = z
On differentiating both sides we have,
cosxdx = dz
ycosx = -2∫zdz + c
ycosx = -2(z²/2) + c
ycosx = -sin²xdx + c
y = secx(-sin²xdx + c)
This is the required solution.
Question 28. (dy/dx) + ycosx = sinx.cosx
Solution:
We have,
(dy/dx) + ycosx = sinx.cosx  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cosx, Q = sinx.cosx
So,
I.F = e^∫Pdx
= e^∫cosxdx
= e^sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^sinx) = ∫(e^sinx)(sinx.cosx)dx + c
Let, sinx = z
Differentiating both sides we get,
cosxdx = dz
y(e^z) = ∫ze^zdz + c
y(e^z) = z∫e^zdz – {(dz/dz)∫e^zdz}dz
y(e^z) = ze^z– ∫e^zdz + c
y(e^z) = e^z(z – 1) + c
y = (z – 1) + ce^-z
y = (sinx – 1) + ce^-sinx
This is the required solution.
Question 29. (1 + x²)(dy/dx) – 2xy = (x²+ 2)(x²+ 1)
Solution:
We have,
(1 + x²)(dy/dx) – 2xy = (x²+ 2)(x²+ 1)
(dy/dx) – 2xy/(1 + x²) = (x²+ 2)  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -2x/(1 + x²), Q = (x²+ 1)
So,
I.F = e^∫Pdx
= $e^{-∫\frac{2x}{1+x^2}dx}$
$=e^{-log|x^2+1|}$
= 1/(x²+1)
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y.[1/(x²+ 1)] = ∫[(x²+ 2)/(x²+ 1)]dx + c
y/(x²+ 1) = ∫[1 + 1/(x²+ 1)]dx + c
y/(x²+ 1) = x + tan^-1x + c
y = (x²+ 1)(x + tan^-1x + c)
This is the required solution.
Question 30. sinx(dy/dx) + ycosx = 2sin²xcosx
Solution:
We have,
sinx(dy/dx) + ycosx = 2sin²xcosx
(dy/dx) + ycotx = 2sinx.cosx  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cotx, Q = 2sinx.cosx
So,
I.F = e^∫Pdx
= e^∫cotxdx
= e^log|sinx|
= sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(sinx) = ∫(2sinx.cosx)sinxdx + c
Let, sinx = z
On differentiating both sides we have,
cosxdx = dz
y.z = 2∫z²+ c
y.z = (2/3)z³+ c
y.sinx = (2/3)sin³x + c
This is the required solution.
Question 31. (x²– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)
Solution:
We have,
(x²– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)
(dy/dx) + 2(x + 2)y/(x²– 1) = 2(x + 1)/(x² – 1)
(dy/dx) + 2(x + 2)y/(x²– 1) = 2/(x – 1)   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2(x + 2)/(x²– 1), Q = 2/(x – 1)
So,
I.F = e^∫Pdx
= $e^{∫\frac{2(x+2)}{x^2-1}dx}$
$=e^{∫\frac{2x}{x^2-1}dx+∫\frac{4}{x^2-1}dx}$
$=e^{log|x^2-1|+\frac{4}{2}log|\frac{x-1}{x+1}|}$
$=e^{log(|x^2-1||\frac{x-1}{x+1}|^2)}$
$=e^{log|\frac{(x-1)^3}{x+1}|}$
= (x – 1)³/(x + 1)
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y.[(x – 1)³/(x + 1)] = ∫[(x – 1)³/(x + 1)[{2/(x – 1)]dx + c
$\frac{(x-1)^3}{(x+1)}y=2∫\frac{(x-1)^2}{(x+1)}+c$
$\frac{(x-1)^3}{(x+1)}y=2∫\frac{(x+1)^2-4x}{(x+1)}+c$
$\frac{(x-1)^3}{(x+1)}y=2∫[(x+1)-4\frac{x}{(x+1)}]+c$
$\frac{(x-1)^3}{(x+1)}y=∫[2x+2-8\frac{x}{x+1}]dx$
y.[(x – 1)³/(x + 1)] = (x²– 6x + 8log|x + 1|) + c
$y=\frac{x+1}{(x-1)^3}(x^2-6x+8log|x+1|+c)$
This is the required solution.
Question 32. (dy/dx) + (2y/x) = cosx
Solution:
We have,
(dy/dx) + (2y/x) = cosx   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2/x, Q = cosx
So,
I.F = e^∫Pdx
= e^∫(2/x)dx
= e^2log|x|
= x²
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x²) = ∫(x²).(cosx)dx + c
x²(y) = x²∫cosxdx – ∫{(d/dx)x²∫cosxdx}dx + c
x²y = x²sinx – 2∫xsinxdx + c
x²y = x²sinx – 2x∫sinxdx + 2∫{(dx/dx)∫sinxdx}dx + c
x²y = x²sinx + 2xcosx – 2∫cosxdx + c
x²y = x²sinx + 2xcosx – 2sinx + c
This is the required solution.
Question 33. (dy/dx) – y = xe^x
Solution:
We have,
(dy/dx) – y = xe^x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1, Q = xe^x
So,
I.F = e^∫Pdx
= e^∫-dx
= e^-x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^-x) = ∫(e^-x)(xe^x)dx + c
ye^-x= ∫xdx + c
ye^-x= (x²/2) + c
y = [(x²/2) + c].e^x
This is the required solution.
Question 34. (dy/dx) + 2y = xe^4x
Solution:
We have,
(dy/dx) + 2y = xe^4x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = xe^4x
So,
I.F = e^∫Pdx
= e^2∫dx
= e^2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = ∫(e^2x).(xe^4x)dx + c
y(e^2x) = ∫xe^6xdx + c
y(e^2x) = x∫e^6xdx – ∫{(dx/dx)∫e^6xdx}dx + c
e^2xy = (xe^6x)/6 – ∫(e^6x/6)dx + c
e^2xy = (xe^6x)/6 – e^6x/36 + c
y = (xe^4x)/6 – e^4x/36 + ce^-2x
This is the required solution.
Question 35. (x + 2y²)(dy/dx) = y, given that when x = 2, y = 1
Solution:
We have,
(x + 2y²)(dy/dx) = y
(dx/dy) = (x + 2y²)/y
(dx/dy) – (x/y) = 2y   ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = 1/y, Q = 2y
So,
I.F = e^∫Pdy
= e^∫-dy/y
= e^-log|y|
= 1/y
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(1/y) = ∫(1/y)(2y)dy + c
(x/y) = 2∫dy + c
(x/y) = 2y + c
x = 2y²+ cy
Given that when x = 2, y = 1
2 = 2 + c
c = 0
x = 2y²
This is the required solution.
Question 36(i). Find one-parameter families of solution curves of the following differential equation (dy/dx) + 3y = e^mx, m is a given real number
Solution:
We have,
(dy/dx) + 3y = e^mx      ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 3, Q = e^mx
So,
I.F = e^∫Pdx
= e^∫3dx
= e^3x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^3x) = ∫(e^3x).(e^mx)dx + c
y(e^3x) = ∫e^(3+m)xdx + c
y(e^3x) = e^(m+3)x/(m + 3) + c
y = e^mx/(m + 3) + c
This is the required solution.
Question 36(ii). Find one-parameter families of solution curves of the following differential equation (dy/dx) – y = cos2x
Solution:
We have,
(dy/dx) – y = cos2x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1, Q = cos2x
So,
I.F = e^∫Pdx
= e^-∫dx
= e^-x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^-x) = ∫(e^-x).(cos2x)dx + c
y(e^-x) = ∫e^-xcos2xdx + c
Let,
A = ∫e^-xcos2xdx
= e^-x∫cos2xdx – {(d/dx)e^-x∫cos2xdx}dx
= (e^-x/2)sin2x + ∫(e^-x/2)sin2xdx
= $e^{-x}sin2x+\frac{1}{2}e^{-x}∫sin2x-\frac{1}{2}∫[\frac{d}{dx}e^{-x}∫sin2xdx]$
$=\frac{1}{2}e^{-x}sin2x-\frac{1}{4}e^{-x}cos2x-\frac{1}{4}∫e^{-x}cos2xdx$
$A=e^{-x}sin2x-\frac{1}{4}e^{-x}cos2x-\frac{1}{4}A$
(5/4)A = (e^-x/2)(2sin2x – cos2x)
$A=\frac{2}{5}e^{-x}(2sin2x-cos2x)$
$ye^{-x}=\frac{2}{5}e^{-x}(2sin2x-cos2x)+c$
$y=\frac{2}{5}(2sin2x-cos2x)+ce^x$
This is the required solution.
Question 36(iii). Find one-parameter families of solution curves of the following differential equation x(dy/dx) – y = (x + 1)e^-x
Solution:
We have,
x(dy/dx) – y = (x + 1)e^-x
(dy/dx) – y/x = [(x + 1)/x]e^-x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1/x, Q = [(x + 1)/x]e^-x
So,
I.F = e^∫Pdx
= e^-∫dx/x
= e^-log|x|
= 1/x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/x) = ∫[(x+1)/x]e^-x(1/x)dx + c
y/x = ∫[1/x+1/x²]e^-xdx + c
Let, (1/x)e^-x= z
On differentiating both sides we have
-[1/x + 1/x2]e^-xdx = dz
y/x = -∫dz + c
y/x = -z + c
y/x = -(e^-x/x) + c
y = -e^-x+ cx
This is the required solution.
Question 36(iv). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + y = x⁴
Solution:
We have,
x(dy/dx) + y = x⁴
(dy/dx) + y/x = x³     ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/x, Q = x³
So,
I.F = e^∫Pdx
= e^∫dx/x
= e^log|x|
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x) = ∫(x)(x³)dx + c
xy = ∫x⁴+ c
xy = (x⁵/5) + c
y = (x⁴/5) + c/x
This is the required solution.
Question 36(v). Find one-parameter families of solution curves of the following differential equation (xlogx)(dy/dx) + y = logx
Solution:
We have,
(xlogx)(dy/dx) + y = logx
(dy/dx) + y/xlogx = 1/x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/xlogx, Q = 1/x
So,
I.F = e^∫Pdx
= e^∫dx/xlogx
Let, logx = z
On differentiating both sides we have
dx/x = dz
= e^∫dz/z
= e^log|z|
= z
=l ogx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(logx) = ∫(1/x)(logx)dx + c
y(logx) = ∫zdz + c (Let, logx=z and differentiating both sides)
y(logx) = (z²/2) + c
y(logx) = (logx)²/2 + c
y = logx/2 + c/logx
This is the required solution.
Question 36(vi). Find one-parameter families of solution curves of the following differential equation (dy/dx) – 2xy/(1 + x²) = x²+ 2
Solution:
We have,
(dy/dx) – 2xy/(1 + x²) = x²+ 2   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -2x/(1 + x²), Q = x²+ 2
So,
I.F = e^∫Pdx
= e^{-∫2xdx/(1+x2)}
= e^-log|1+x2|
= 1/(1+x²)
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y[1/(1 + x²)] = ∫[1/(1 + x²)](x²+ 2)dx + c
y/(1 + x²) = ∫[(x²+ 2)/(x²+ 1)]dx + c
$\frac{y}{x^2+1}=∫\frac{x^2+1+1}{x^2+1}dx+c$
y/(1 + x²) = ∫dx + ∫dx/(x²+ 1) + c
y/(x²+ 1) = x + tan^-1x + c
y = (x + tan^-1x + c)(x²+ 1)
This is the required solution.
Question 36(vii). Find one-parameter families of solution curves of the following differential equation (dy/dx) + ycosx = e^sinxcosx
Solution:
We have,
(dy/dx) + ycosx = e^sinxcosx    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cosx, Q = e^sinxcosx
So,
I.F = e^∫Pdx
= e∫cosxdx
= e^sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y[(e^sinx) = ∫(e^sinx)(e^sinxcosx)dx + c
Let, sinx = z
ON differentiating both sides we have,
cosxdx = dz
ye^z= ∫e^2zdz + c
ye^z= (e^2z/2) + c
y = ez/2 + ce – z
y = (e^sinx/2) + ce^-sinx
This is the required solution.
Question 36(viii). Find one-parameter families of solution curves of the following differential equation (x + y)(dy/dx) = 1
Solution:
We have,
(x + y)(dy/dx) = 1
(dy/dx) = 1/(x + y)
(dx/dy) = (x + y)
(dx/dy) – x = y    ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = -1, Q = y
So,
I.F = e^∫Pdy
= e^-∫dy
= e^-y
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(e^-y) = ∫(e^-y)(y)dy + c
xe^-y= y∫e^-ydy – ∫{(dy/dy)∫e^-ydy}dy + c
xe^-y= -ye^-y+ ∫e^-y+ c
xe^-y= -ye^-y– e^-y+ c
xe^-y+ ye^-y+ e^-y= c
e^-y(x + y + 1) = c
(x + y + 1) = ce^y
This is the required solution.
Question 36(ix). Find one-parameter families of solution curves of the following differential equation cos²x(dy/dx) = (tanx – y)
Solution:
We have,
cos²x(dy/dx) = (tanx – y)
(dy/dx) = (tanx – y)/cos²x
(dy/dx) = tanx.sec²x – ysec²x
(dy/dx) + ysec²x = tanx.sec²x    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = sec²x, Q = tanx.sec²x
So,
I.F = e∫Pdx
$=e^{∫sec^2xdx}$
= e^tanx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^tanx) = ∫(e^tanx)(tanx.sec²x)dx + c
Let, tanx = z
On differentiating both sides we have,
sec²xdx = dz
y(e^z) = ∫ze^zdz + c
y(e^z) = z∫e^zdz – ∫{(dz/dz)∫e^zdz}dz
y(e^z) = ze^z– ∫e^zdz + c
y(e^z) = ze^z– e^z+ c
y = (z – 1) + c.e^-z
y = (tanx – 1) + c.e^-tanx
This is the required solution.
Question 36(x). Find one-parameter families of solution curves of the following differential equation e^-ysec²ydy = dx + xdy
Solution:
We have,
dx + xdy = e^-ysec²ydy
(x – e^-ysec²y)dy = -dx
(dx/dy) = (e^-ysec²y – x)
(dx/dy) + x = e^-ysec²y    ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = 1, Q = e^-ysec²y
So, I.F = e^∫Pdy
= e^∫dy
= e^y
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(e^y) = ∫e^-ysce²ye^ydy + c
x(e^y) = ∫sec²ydy + c
x(e^y) =tany + c
x = (tan y + c)e^-y
This is the required solution.
Question 36(xi). Find one-parameter families of solution curves of the following differential equation (xlogx)(dy/dx) + y = 2logx
Solution:
We have,
(xlogx)(dy/dx) + y = 2logx
(dy/dx) + y/xlogx = 2/x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/xlogx, Q = 2/x
So,
I.F = e^∫Pdx
= e^∫dx/xlogx
Let, logx = z
On differentiating both sides we have
dx/x = dz
= e^∫dz/z
= e^log|z|
= z
= logx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(logx) = ∫(2/x)(logx)dx + c
y(logx) = 2∫zdz + c (Let, logx = z and differentiating both sides)
y(logx) = 2(z²/2) + c
y(logx) = (logx)²+ c
y = logx + c/logx
This is the required solution.
Question 36(xii). Find one-parameter families of solution curves of the following differential equation x(dy/dx) + 2y = x²logx
Solution:
We have,
x(dy/dx) + 2y = x²logx
(dy/dx) + 2y/x = xlogx   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2/x, Q = xlogx
So,
I.F = e∫Pdx
= e^2∫dx/x
= e^2logx
= x²
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x²) = ∫(x²)(xlogx)dx + c
x²y = ∫x³logxdx + c
x²y = logx∫x³dx + ∫{(d/dx)logx∫x³dx}dx + c
x²y = (1/4)x⁴logx – (1/4)∫x³dx + c
x²y = (1/4)x⁴logx – (1/16)x⁴+ c
y = (x²/16)(4logx – 1) + c/x²
This is the required solution.
Question 37. Solve the following using the initial value problem:-
(i). y’ + y = e^x, y(0) = (1/2)
Solution:
We have,
y’ + y = e^x
dy/dx + y = e^x    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1, Q = e^x
So, I.F = e^∫Pdx
= e^∫dx
= e^x
The solution of differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^x) = ∫e^x.e^xdx + c
y(e^x) = (1/2)e^2x+ c
At t = 0, y = (1/2)
(1/2)e⁰= (1/2)e⁰+ c
c = 0
y(e^x) = (1/2)e^2x
y = (e^x/2)
This is the required solution.
(ii). x(dy/dx) – y = logx, y(1) = 0
Solution:
We have,
x(dy/dx) – y = logx
(dy/dx) – y/x = logx/x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1/x, Q = logx/x
So,
I.F = e^∫Pdx
= e^-∫dx/x
= e^-log|x|
= 1/x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/x) = ∫(1/x)(logx/x)dx + c
(y/x) = ∫(logx/x²)dx + c
(y/x) = logx∫(dx/x²) – ∫{(d/dx)logx∫(dx/x²)}dx + c
(y/x) = -(logx/x) + ∫(dx/x²) + c
(y/x) = -(logx/x) – (1/x) + c
At x = 1, y = 0
0 = -0 – 1 + c
c = 1
(y/x) = -(logx/x) – (1/x) + 1
y = x – 1 – logx
This is the required solution.
(iii). (dy/dx) + 2y = e^-2xsinx, y(0) = 0
Solution:
We have,
(dy/dx) + 2y = e^-2xsinx    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = e^-2xsinx
So,
I.F = e^∫Pdx
= e^∫2dx
= e^2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^2x) = ∫e^-2xsinx.(e^2x)dx + c
y(e^2x) = ∫sinxdx + c
y(e^2x) = -cosx + c
At x = 0, y = 0
0 = -1 + c
c = 1
y(e^2x) = 1 – cosx
This is the required solution.
(iv). x(dy/dx) – y = (x + 1)e^-x, y(1) = 0
Solution:
We have,
x(dy/dx) – y = (x + 1)e^-x
(dy/dx) – (y/x) = [(x + 1)/x]e^-x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -(1/x), Q = [(x + 1)/x]e^-x
So,
I.F = e^∫Pdx
= e^-∫(dx/x)
= e^-log(x)
= e^log(1/x)
= 1/x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
$y(\frac{1}{x})=∫(\frac{x+1}{x})e^{-x}.(\frac{dx}{x})+c$
(y/x) = ∫[(1/x) + (1/x²)]e^-x+ c
Since, -∫[f(x) + f'(x)]e^-xdx = f(x)e^-x+ c
(y/x) = -e^-x/x + c
At x = 1, y = 0
0 = -e^-1+ c
c = e^-1
(y/x) = -e^-x/x + e^-1
y = xe^-1– e^-x
This is the required solution.
(v). (1 + y²)dx + (x – $e^{-tan^{-1}y}$ )dx = 0, y(0) = 0
Solution:
We have,
(1 + y²)(dx/dy) + x = $e^{-tan^{-1}y}$
(dy/dx) + [1/(y²+ 1)]x = $e^{-tan^{-1}y}$ /(y²+ 1)   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Px = Q
Where, P = 1/(y²+ 1), Q = $e^{-tan^{-1}y}$ /(y²+ 1)
So,
I.F = e^∫Pdy
$=e^{∫\frac{1}{y^2+1}dy}$
= e^tan-1y
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x( $e^{-tan^{-1}y}$ ) = ∫[ $e^{-tan^{-1}y}$ /(y²+ 1)] $e^{-tan^{-1}y}$ dx + c
x( $e^{-tan^{-1}y}$ ) = ∫dy/(1 + y²) + c
x(e^tan-1y) = tan^-1y + c
At x = 0, y = 0
0*e⁰= 0 + c
c = 0
x( $e^{-tan^{-1}y}$ ) = tan^-1y
This is the required solution.
(vi). (dy/dx) + ytanx = x²tanx + 2x, y(0) = 1
Solution:
We have,
(dy/dx) + ytanx = x²tanx + 2x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = tanx, Q = x²tanx+2x
So,
I.F = e^∫Pdx
= e^∫tanxdx
= e^log|secx|
= secx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(secx) = ∫(x²tanx + 2x)secxdx + c
y(secx) = ∫(x²tanxsecx + 2xsecx)dx + c
y(secx) = ∫x²tanxsecxdx + 2∫xsecxdx + c
y(secx) = ∫x²tanxsecxdx + 2secx∫xdx – 2∫{(d/dx)secx∫xdx}dx + c
y(secx) = ∫x²tanxsecxdx + x².secx – ∫x²tanxsecxdx + c
y(secx) = x²,(secx)+c
At, x = 0, y = 1
1 = 0 + c
c = 1
y = x²+ cosx
This is the required solution.
(vii). x(dy/dx) + y = xcosx + sinx, y(π/2) = 1
Solution:
We have,
x(dy/dx) + y = xcosx + sinx
(dy/dx) + (y/x) = cosx + sinx/x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/x, Q = cosx + sinx/x
So,
I.F = e^∫Pdx
= e^∫dx/x
= e^log|x|
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x) = ∫(cosx + sinx/x)(x)xdx + c
y(x) = ∫xcosxdx + ∫sinxdx + c
xy = x∫cosxdx – ∫{(dx/dx)∫cosxdx}dx – cosx + c
xy = xsinx – ∫sinxdx – cosx + c
xy = xsinx + cosx – cosx + c
xy = xsinx + c
At x = π/2, y = 1
π/2 = π/2sin(π/2) + c
c = 0
y = sinx
This is the required solution.
(viii). (dy/dx) + ycotx = 4xcosecx, y(π/2) = 0
Solution:
We have,
(dy/dx) + ycotx = 4xcosecx   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cotx, Q = 4xcosecx
So,
I.F = e^∫Pdx
= e^∫cotx.dx
= e^log|sinx|
= sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(sinx) = 4∫(xcosecx)(sinx)xdx + c
y(sinx) = 4∫xdx + c
y(sinx) = 4(x²/2) + c
y(sinx) = 2x²+ c
At x = π/2, y = 0,
0 = 2(π/2)²+ c
c = -π²/2
y(sinx) = 2x²– π²/2
This is the required solution.
(ix). (dy/dx) + 2ytanx = sinx, y = 0 when x = π/3
Solution:
We have,
(dy/dx) + 2ytanx = sinx   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2tanx, Q = sinx
So, I.F = e^∫Pdx
= e^∫2tanxdx
= e^2log|secx|
= sec²x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y.sec²x = ∫sinx.sec²xdx + c
y.sec²x = ∫tanx.secxdx + c
y.sec²x = secx+c
At x = π/3, y = 0,
0 = sec²(π/3) + c
c = -2
y.sec²x = secx – 2
y = cosx – 2cos²x
This is the required solution.
(x). (dy/dx) – 3ycotx = sin2x, y = 2 when x = π/2
Solution:
We have,
(dy/dx) – 3ycotx = sin2x    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -3cotx, Q = sin2x
So, I.F = e^∫Pdx
= e^∫-3cotxdx
= e^-3log|sinx|
= cosec³x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y.(cosec³x) = 2∫(cosec³x).(sin2x)dx + c
y.(cosec³x) = 2∫cotx.cosecxdx + c
y.(cosec³x) = -2cosesx +c
y = -2sin²x + c.sin³x
At x = π/2, y = 2.
2 = -2sin²(π/2) + c.sin³(π/2)
c = 4
y = c.sin³x – 2sin²x
This is the required solution.
(xi). (dy/dx) + ycotx = 2cosx, y(π/2) = 0
Solution:
We have,
(dy/dx) + ycotx = 2cosx    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cotx, Q = 2cosx
So, I.F = e^∫Pdx
= e^∫cotxdx
= e^log|sinx|
= sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y.(sinx) = ∫(sinx).(2cosx)dx + c
y.(sinx) = 2∫sinx.cosxdx + c
y.(sinx) = ∫sin2xdx + c
y.(sinx) = -(cos2x/2) + c
At x = π/2, y = 0
0 = -cos(π)/2 + c
c = -(1/2)
y.(sinx) = -(cos2x/2) – (1/2)
2y(sinx) = -(1 + cos2x)
2y(sinx) = -2cos²x
y = -cotx.cosx
This is the required solution.
(xii). dy = cosx(2 – ycosecx)dx,
Solution:
We have,
dy = cosx(2 – ycosecx)dx
(dy/dx) = -ycotx + 2cosx
(dy/dx) + ycotx = 2cosx    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cotx, Q = 2cosx
So,
I.F = e^∫Pdx
= e^∫cotxdx
= e^log|sinx|
= sinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(sinx) = 2∫cosx.(sinx)dx + c
ysinx = ∫2cosx.sinxdx + c
ysinx = ∫sin2x + c
ysinx = -(cos2x/2) + c
This is the required solution.
Question 38. x(dy/dx) + 2y = x²
Solution:
We have,
x(dy/dx) + 2y = x²
(dy/dx) + 2y/x = x    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2/x, Q = x
So, I.F = e^∫Pdx
= e^2∫dx/x
= e^2logx
= x²
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
yx²= ∫x2.xdx + c
yx²= ∫x³dx + c
x²y = (x⁴/4) + c
y = (x1/4) + c.x^-2
This is the required solution.
Question 39. (dy/dx) – y = cosx
Solution:
We have,
(dy/dx) – y = cosx   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1, Q = cosx
So, I.F = e^∫Pdx
= e^-∫dx
= e^-x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e^-x) = ∫cosx.e^-xdx + c
Let, I = ∫cosx.e^-xdx
I = e^-x∫cosxdx – ∫{(d/dx)e^-x∫cosxdx}dx
I = e^-xsinx + ∫e^-xsinxdx
I = e^-xsinx + e^-x∫sinxdx-∫{(d/dx)e^-x∫sinxdx}dx
I = e^-xsinx – e^-xcosx-∫e^-xcosxdx
2I = e^-x(sinx – cosx)
I = (e^-x/2)(sinx – cosx)
y(e^-x) = (e^-x/2)(sinx – cosx) + c
y = (1/2)(sinx-cosx) + ce^x
This is the required solution.
Question 40. (y + 3x²)(dx/dy) = x
Solution:
We have,
(y + 3x²)(dx/dy) = x
(dy/dx) = (y + 3x²)/x
(dy/dx) – y/x = 3x    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1/x, Q = 3x
So, I.F = e^∫Pdx
= e^-∫dx/x
= e^-logx
= 1/x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/x) = 3∫x.(1/x)dx + c
y(1/x) = 3∫dx + c
y/x = 3x + c
This is the required solution.
Question 41. Find a particular solution of the differential equation (dx/dy) + xcoty = y²coty + 2y, given that x = 0, when y = π/2
Solution:
We have,
(dx/dy) + xcoty = y²coty + 2y   ………..(i)
The given equation is a linear differential equation of the form
(dx/dy) + Px = Q
Where, P = coty, Q = y²coty + 2y
So,
I.F = e^∫Pdy
= e^∫cotydy
= e^log|siny|
= siny
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(siny) = ∫(y²coty + 2y)sinydy + c
x(siny) = ∫(y²cosy + 2xsiny)dy + c
x(siny) = y²∫cosydx – {(d/dy)y²∫cosydy}dy + ∫2ysinydy + c
x(siny) = y²siny – ∫2ysinydy + ∫2ysinydy + c
x(siny) = y²siny + c
At x = 0, y = π/2
0 = (π/2)²sin(π/2) + c
c = -π²/4
x(siny) = y²siny – π²/4
This is the required solution.

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RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations

Solve the following differential equations:

Question 1. x2dy + y(x + y)dy = 0

Question 2. (dy/dx) = (y – x)/(y + x)

Question 3. (dy/dx) = (y2 – x2)/2yx

Question 4. x(dy/dx) = (x + y)

Question 5. (x2 – y2)dx – 2xydy = 0

Question 6. (dy/dx) = (x + y)/(x – y)

Question 7. 2xy(dy/dx) = (x2 + y2)

Question 8. x2(dy/dx) = x2 – 2y2 + xy

Question 9. xy(dy/dx) = x2 – y2

Question 10. yex/ydx = (xex/y + y)dy

Question 11. x2(dy/dx) = x2 + xy + y2

Question 12. (y2 – 2xy)dx = (x2 – 2xy)dy

Question 13. 2xydx + (x2 + 2y2)dy = 0

Question 14. 3x2dy = (3xy + y2)dx

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

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

Question 17.

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

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

Question 20. y2dx + (x2 – xy + y2)dy = 0

Question 21. [x√(x2 + y2) – y2]dx + xydy = 0

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

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

Question 24. xylog(x/y)dx + {y2 – x2log(x/y)}dy = 0

Question 25. (1 + ex/y)dx + ex/y(1 – x/y)dy = 0

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

Question 27. (x2 – 2xy)dy + (x2 – 3xy + 2y2)dx = 0

Question 28. x(dy/dx) = y – xcos2(y/x)

Question 29. x(dy/dx) – y = 2√(y2 – x2)

Question 30. xcos(y/x)(ydx + xdy) = ysin(y/x)(xdy – ydx)

Question 31. (x2 + 3xy + y2)dx – x2dy = 0

Question 32. (x – y)(dy/dx) = (x + 2y)

Question 33. (2x2y + y3)dx + (xy2 – 3x2)dy = 0

Question 34. x(dy/dx) – y + xsin(y/x) = 0

Question 35. ydx + {xlog(y/x)}dy – 2xydy = 0

Question 36(i). (x2 + y2)dx = 2xydy, y(1) = 0

Question 36(ii). xex/y – y + x(dy/dx) = 0, y(e) = 0

Question 36(iii). (dy/dx) – (y/x) + cosec(y/x) = 0, y(1) = 0

Question 36(iv). (xy – y2)dx – x2dy = 0, y(1) = 1

Question 36(v). (dy/dx) = [y(x + 2y)]/[x(2x + y)]

Question 36(vi). (y4 – 2x3y)dx + (x4 – 2xy3)dy = 0, y(1) = 0

Question 36(vii). x(x2 + 3y2)dx + y(y2 + 3x2)dy = 0, y(1) = 1

Question 36(viii). {xsin2(y/x) – y}dx + xdy = 0, y(1) = π/4

Question 36(ix). x(dy/dx) – y + xsin(y/x) = 0, y(2) = π

Question 37. xcos(y/x)(dy/dx) = ycos(y/x) + x, When x = 1, y = π/4

Question 38. (x – y)(dy/dx) = (x + 2y), when x = 1,y = 0

Question 39. (dy/dx) = xy/(x2 + y2)

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Question 1. x²dy + y(x + y)dy = 0

Question 3. (dy/dx) = (y²– x²)/2yx

Question 5. (x²– y²)dx – 2xydy = 0

Question 7. 2xy(dy/dx) = (x²+ y²)

Question 8. x²(dy/dx) = x²– 2y²+ xy

Question 9. xy(dy/dx) = x²– y²

Question 10. ye^x/ydx = (xe^x/y+ y)dy

Question 11. x²(dy/dx) = x²+ xy + y²

Question 12. (y²– 2xy)dx = (x²– 2xy)dy

Question 13. 2xydx + (x²+ 2y²)dy = 0

Question 14. 3x²dy = (3xy + y²)dx

Question 17. $\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}$

Question 20. y²dx + (x²– xy + y²)dy = 0

Question 21. [x√(x²+ y²) – y²]dx + xydy = 0

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

Question 24. xylog(x/y)dx + {y²– x²log(x/y)}dy = 0

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

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

Question 27. (x²– 2xy)dy + (x²– 3xy + 2y²)dx = 0

Question 28. x(dy/dx) = y – xcos²(y/x)

Question 29. x(dy/dx) – y = 2√(y²– x²)

Question 31. (x²+ 3xy + y²)dx – x²dy = 0

Question 33. (2x²y + y³)dx + (xy²– 3x²)dy = 0

Question 36(i). (x²+ y²)dx = 2xydy, y(1) = 0

Question 36(ii). xe^x/y– y + x(dy/dx) = 0, y(e) = 0

Question 36(iv). (xy – y²)dx – x²dy = 0, y(1) = 1

Question 36(vi). (y⁴– 2x³y)dx + (x⁴– 2xy³)dy = 0, y(1) = 0

Question 36(vii). x(x²+ 3y²)dx + y(y²+ 3x²)dy = 0, y(1) = 1

Question 36(viii). {xsin²(y/x) – y}dx + xdy = 0, y(1) = π/4

Question 39. (dy/dx) = xy/(x²+ y²)