RD Sharma Class 12 Ex 22.9 Solutions Chapter 22 Differential Equations

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter22
Exercise22.9
CategoryRD 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}

yx= (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 – v– v)/(v + 1)

x(dv/dx) = -(v+ 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|v+ 1| + tan-1(v) = log(c/x)

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

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

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

Question 3. (dy/dx) = (y– x2)/2yx

Solution:

We have,

(dy/dx) = (y– 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) = (v2x– x2)/2vx2

v + x(dv/dx) = (v– 1)/2v

x(dv/dx) = [(v– 1)/2v] – v

x(dv/dx) = (v– 1 – 2v2)/2v

x(dv/dx) = -(v+ 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/x+ 1 = |c/x|

(x+ 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. (x– y2)dx – 2xydy = 0

Solution:

We have,

(x– y2)dx – 2xydy = 0

(dy/dx) = (x– 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) = (x– 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)+ log(c)

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

(x– 3y2)/x= (c/x3)

x(x– 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/(v+ 1) – ∫vdv/(v+ 1) = ∫(dx/x)

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

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

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

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

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

Solution:

We have,

2xy(dy/dx) = (x+ y2)

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

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

Question 8. x2(dy/dx) = x– 2y+ xy

Solution: 

We have,

x2(dy/dx) = x– 2y+ xy

(dy/dx) = (x– 2y+ 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) = (x– 2v2x+ xvx)/2xvx

v + x(dv/dx) = (1 – 2v+ v)/x2

x(dv/dx) = (1 – 2v+ 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) = x– y2

Solution:

We have,

xy(dy/dx) = x– y2

(dy/dx) = (x– 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) = (x– 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)/x= c/x4

x2(x– 2y2) = c  (Where ‘c’ is integration constant)

Question 10. yex/ydx = (xex/y + y)dy

Solution:

We have,

yex/ydx = (xex/y + y)dy

(dy/dx) = (xex/y + y)/yex/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) = (vyevy/y + y)/yevy/y

v + y(dv/dy) = (ve+ 1)/ev

y(dv/dy) = [(ve+ 1)/ev] – v

y(dv/dy) = (ve+ 1 – vev)/ev

y(dv/dy) = (1/ev)

evdv = (dy/y)

On integrating both sides,

∫evdv = ∫(dy/y)

e= log(y) + log(c)

ex/y = log(y) + log(c)  (Where ‘c’ is integration constant)

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

Solution:

We have,

x2(dy/dx) = x+ xy + y2

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

Solution:

We have,

(y– 2xy)dx = (x– 2xy)dy

(dy/dx) = (y– 2xy)/(x– 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) = (v2x– 2xvx)/(x– 2xvx)

v + x(dv/dx) = (v– 2v)/(1 – 2v)

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

x(dv/dx) = 3(v– 1)/(1 – 2v)

-(2v – 1)dv/(v– v) = 3(dx/x)

On integrating both sides,

-∫(2v – 1)dv/(v– v) = 3∫(dx/x)

-log|v– v| = 3log|x| – log|c|

log|v– v| = log|c/x3|

(y2/x– y/x) = (c/x3)

(y– xy) = c/x

x(y– xy) = c (Where ‘c’ is integration constant)

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

Solution:

We have,

2xydx + (x+ 2y2)dy = 0

dy/dx = -(2xy)/(x+ 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)/(x+ 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)= (c/x)3

(3yx+ 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 + v– 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 – 2v– v)/(2v + 1)

(2v + 1)dv/(2v+ v – 1) = -(dx/x)

On integrating both sides,

∫(2v + 1)dv/(2v+ 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|+ 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/x– 1)

v + x(dv/dx) = v – √(v– 1)

x(dv/dx) = -√(v– 1)

dv/√(v– 1) = -(dx/x)

On integrating both sides,

∫dv/√(v– 1) = -∫(dx/x)

log|v + √(v– 1)| = -log|x| + log|c|

|v + √(v– 1)| = (c/x)

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

y + √(y– 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 + (x– xy + y2)dy = 0

Solution:

We have,

y2dx + (x– xy + y2)dy = 0

(dy/dx) = -(y2)/(x– 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)/(x– 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√(x+ y2) – y2]dx + xydy = 0

Solution:

We have,

[x√(x+ y2) – y2]dx + xydy = 0

dy/dx = -[x√(x+ y2) – y2]/xy

dy/dx = [y– x√(x+ 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) = [v2x– x√(x+ v2x2)]/xvx

v + x(dv/dx) = [v– √(1 + v2)]/v

x(dv/dx) = [v– √(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 + v= z

On differentiating both sides,

2vdv = dz

(1/2)∫dz/√z = -∫(dx/x)

√z = -log|x| + log|c|

√(1 + v2) = log|c/x|

√(x+ y2)/x = log|c/x|

√(x+ 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 + {y– x2log(x/y)}dy = 0

Solution:

We have,

xylog(x/y)dx + {y– 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 + ex/y)dx + ex/y(1 – x/y)dy = 0

Solution:

We have,

(1 + ex/y)dx + ex/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) = -[ev(1 – v)/(1 + ev)] – v

y(dv/dy) = (-e+ ve– v – vev)/(1 + ev)

y(dv/dy) = -(v + vev)/(1 + ev)

[(1 + ev)/(v + ev)]dv = -(dy/y)

On integrating both sides,

∫[(1 + ev)/(v + ev)]dv = -∫(dy/y)

log|(v + ev)| = -log(y) + log(c)

log|(v + ev)| = log|c/y|

(x/y) + ex/y = c/y

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

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

Solution:

We have,

(x+ y2)dy/dx = (8x– 3xy + 2y2)

(dy/dx) = (8x– 3xy + 2y2)/(x+ 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) = (8x– 3xvx + 2v2x2)/(x+ v2x2)

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

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

x(dv/dx) = (8 – 4v + 2v– v3)/(1 + v2)

(1 + v2)dv/(8 – 4v + 2v– 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 – Av+ 2B – Bv + 4C + Cv2

(1 + v2) = (C – A)v+ (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.  (x– 2xy)dy + (x– 3xy + 2y2)dx = 0

Solution:

We have,

(x– 2xy)dy + (x– 3xy + 2y2)dx = 0

(dy/dx) = (x– 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) = (x– 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 + 2v– 2v+ 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√(y– x2)

Solution:

We have,

x(dy/dx) – y = 2√(y– x2)

(dy/dx) = [2√(y– 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√(v– 1)

dv/√(v– 1) = 2(dx/x)

On integrating both sides,

∫dv/√(v– 1) = 2∫(dx/x)

log|v + √(v2– 1)| = 2log(x) + log(c)

|v + √(v– 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. (x+ 3xy + y2)dx – x2dy = 0

Solution:

We have,

(x+ 3xy + y2)dx – x2dy = 0

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

On integrating both sides,

∫dv/(1 + v)= ∫(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 + (xy– 3x2)dy = 0

Solution:

We have,

(2x2y + y3)dx + (xy– 3x2)dy = 0

dy/dx = (2x2y + y3)/(3x– 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)/(3x– xv2x2)

v + x(dv/dx) = (2v + v3)/(3 – v3)

x(dv/dx) = [(2v + v3)/(3 – v3)] – v

x(dv/dx) = (2v + v– 3v + v3)/(3 – v3)

(3 – v3)dv/(2v– v) = (dx/x)

On integrating both sides,

∫[(3 – v3)/(2v– 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 – v= A(2v– 1) + (Bv + C)v

3 – v= 2Av– A + Bv+ Cv

3 – v= 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}}

|2y– x2|= 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). (x+ y2)dx = 2xydy, y(1) = 0

Solution:

We have,

 (x+ y2)dx = 2xydy

(dy/dx) = (x+ 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) = (x+ v2x2)/2vx2

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

x(dv/dx) = [(1 + v2)/2v] – v

x(dv/dx) = (1 + v– 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|

|x– y2| = |cx|

At x = 1, y = 0

1 – 0 = c

c = 1

|x– y2| = |x|

(x– y2) = x

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

Solution:

We have,

xex/y – y + x(dy/dx) = 0

(dy/dx) = (y – xex/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 – xev)/x

v + x(dv/dx) = v – ev

x(dv/dx) = v – e– v

x(dv/dx) = -ev

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 – v– 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 + 2v– 2v – v2)/(2 + v)

x(dv/dx) = (v– 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)= v2|xc|

(y – x)3/x= (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). (y– 2x3y)dx + (x– 2xy3)dy = 0, y(1) = 0

Solution:

We have,

(y– 2x3y)dx + (x– 2xy3)dy = 0

dy/dx = (2x3y – y4)/(x– 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)/(x– 2xv3x3)

v + x(dv/dx) = (2v – v4)/(1 – 2v3)

x(dv/dx) = [(2v – v4)/(1 – 2v3)] – v

x(dv/dx) = (2v – v– 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)/(x+ y3) = (1/2)x

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

Solution:

We have,

x(x+ 3y2)dx + y(y+ 3x2)dy = 0

dy/dx = -[x(x+ 3y2)/y(y+ 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 + 3v+ v+ 3v2)/v(v+ 3)

[(v+ 3v)/(1 + 6v+ 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|v+ 6v+ 1| = -log|x|+ log|c|

|v+ 6v+ 1| = |c/x4|

(y+ 6x2y+ x4) = c

At y = 1, x = 1

(1 + 6 + 1) = c

c = 8

(y+ 6x2y+ 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/(x+ y2)

Solution:

We have,

(dy/dx) = xy/(x+ 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/(x+ 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/v– ∫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 = e3x
Solution:

We have,
dy/dx + 2y = e3x  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = e3x
So, I.F = e∫Pdx
= e∫2dx
= e2x
The solution of differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = ∫e3x.e2xdx + c
y(e2x) = (1/5)e5x + c
y = (e3x/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
= e2x
The solution of differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = (5/4)∫e-3x.e2xdx + c
y(e2x) = (5/4)∫e-xdx + c
y = -(5/4)e-3x + ce-2x
This is the required solution.
Question 3. (dy/dx) + 2y = 6ex
Solution:
We have,
(dy/dx) + 2y = 6ex ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = 6ex
So, I.F = e∫Pdx
= e∫2dx
= e2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = ∫6ex.e2xdx + c
y(e2x) = 6∫e3xdx + c
y(e2x) = 2e3x + c
y = 2e+ 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
= ex
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(ex) = ∫e-2x.exdx + c
y(ex) = ∫e-xdx + c
y(ex) = -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)
= elog(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
= e2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = ∫4x.e2xdx + c
y(e2x) = 4x∫e2xdx – 4∫{(dx/dx)∫e2xdx}dx + c
y(e2x) = 2xe2x – 2∫e2xdx + c
y(e2x) = 2xe2x – e2x + c
y = (2x – 1) + ce-2x
This is the required solution.
Question 7. x(dy/dx) + y = xex
Solution:
We have,
x(dy/dx) + y = xex
(dy/dx) + (y/x) = ex  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = (1/x), Q = ex
So,
I.F = e∫Pdx
= e∫(dx/x)
= elog(x)
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x) = ∫x.exdx + c
xy = x∫exdx – {(dx/dx)∫exdx}dx + c
xy = xe– ∫exdx + c
xy = xe– e+ c
y=\frac{x-1}{x}e^x+\frac{c}{x}
This is the required solution.
Question 8. (dy/dx) + [4x/(x+ 1)]y = -1/(x+ 1)2
Solution:
We have,
(dy/dx) + [4x/(x+ 1)]y = -1/(x+ 1)2    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = [4x/(x+ 1)], Q = -1/(x+ 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(x+ 1)= ∫-[1/(x+ 1)2](x+ 1)2dx + c
y(x+ 1)= -∫dx + c
y(x+ 1)= -x + c
y = -x/(x+ 1)+ c/(x+ 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)
= elog(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)ex
Solution:
We have,
x(dy/dx) – y = (x – 1)ex
(dy/dx) – (y/x) = [(x – 1)/x]ex  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -(1/x), Q = [(x – 1)/x]ex
So,
I.F = e∫Pdx
= e-∫(dx/x)
= e-log(x)
= elog(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+ c
Since, ∫[f(x) + f'(x)]exdx = f(x)e+ c
(y/x) = (ex/x) + c
y = e+ 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
= elogx
= 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
= ex
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(ex) = ∫sinx.exdx + c
Let, I = ∫sinx.exdx
I = ex∫sinxdx – ∫{(d/dx)ex∫sinxdx}dx
I = -excos + ∫excosxdx
I = -excosx + ex∫cosxdx – ∫{(d/dx)ex∫cosxdx}dx
I = -excosx + exsinx – ∫exsinxdx
2I = ex(sinx – cosx)
I = (ex/2)(sinx – cosx)
y(ex) = (ex/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
= ex
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(ex) = ∫cosx.exdx + c
Let, I = ∫cosx.exdx
I = ex∫cosxdx – ∫{(d/dx)ex∫cosxdx}dx
I = exsinx – ∫exsinxdx
I = exsinx – ex∫sinxdx + ∫{(d/dx)ex∫sinxdx}dx
I = exsinx + excosx – ∫excosxdx
2I = ex(cosx + sinx)
I = (ex/2)(cosx + sinx)
y(ex) = (ex/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
= e2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = ∫sinx.e2xdx + c
Let, I = ∫sinx.e2xdx
I = e2x∫sinxdx – {(d/dx)e2x∫sinxdx}dx
I = -e2xcosx + 2∫e2xcosdx
I = -e2xcosx + 2e2x∫cosxdx – 2{(d/dx)e2x∫cosxdx}dx
I = -e2xcosx + 2e2xsinx – 4∫e2xsinxdx
5I = e2x(2sinx – cosx)
I = (e2x/5)(2sinx – cosx)
y(e2x) = (e2x/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(ez) = ∫zezdz + c
y(ez) = z∫ezdz – {(dz/dz)∫ezdz}dz
y(ez) = ze– ∫ ezdz + c
y(ez) = ez(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
= elog|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
= elog|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
= elog|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/(x+ 1)]y = e^{tan^{-1}x}/(x+ 1)  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1/(x+ 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}/(x+ 1)]e^{tan^{-1}x}dx + c
Let, tan-1x = z
On differentiating both sides we get
dx/(1 + x2) = dz
ye= ∫e2zdz + c
ye= (e2z/2) + c
y = (ez/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 + 2x+ x2)dx
Solution:
We have,
xdy = (2y + 2x+ x2)dx
(dy/dx) = 2(y/x) + 2x+ x
(dy/dx) – (y/x) = 2x+ x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -(2/x), Q = 2x+ x
So,
I.F = e∫Pdx
= e-2∫(dx/x)
= e-2log(x)
= e2log|1/x|
= 1/x2
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(1/x2) = ∫(1/x2).(2x+ x)dx + c
(y/x2) = ∫[2x + (1/x)]dx + c
(y/x2) = x+ log|x| + c
y = x+ 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= ∫e2zdz + c
xez = (e2z/2) + c
x = ez/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= -∫zezdz + c
xe= -z∫ezdz + ∫{(dz/dz)∫ezdz}dz + c
xe= -ze+ ∫ezdz + c
xe= -ze+ e+ c
x = (1 – z) + ce – z
x = [1 + (1/y)] + ce1/y
x = (y + 1)/y + ce1/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
= e2log|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 = 2y+ 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
= ey
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(ey) = ∫e-ysce2yeydy + c
x(ey) = ∫sec2ydy + c
x(ey) = 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
= esinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(esinx) = ∫(esinx)(sinx.cosx)dx + c
Let, sinx = z
Differentiating both sides we get,
cosxdx = dz
y(ez) = ∫zezdz + c
y(ez) = z∫ezdz – {(dz/dz)∫ezdz}dz
y(ez) = ze– ∫ezdz + c
y(ez) = ez(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 = (x+ 2)(x+ 1)
Solution:
We have,
(1 + x2)(dy/dx) – 2xy = (x+ 2)(x+ 1)
(dy/dx) – 2xy/(1 + x2) = (x+ 2)  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -2x/(1 + x2), Q = (x+ 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/(x+ 1)] = ∫[(x+ 2)/(x+ 1)]dx + c
y/(x+ 1) = ∫[1 + 1/(x+ 1)]dx + c
y/(x+ 1) = x + tan-1x + c
y = (x+ 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
= elog|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∫z+ c
y.z = (2/3)z+ c
y.sinx = (2/3)sin3x + c
This is the required solution.
Question 31. (x– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)
Solution:
We have,
(x– 1)(dy/dx) + 2(x + 2)y = 2(x + 1)
(dy/dx) + 2(x + 2)y/(x– 1) = 2(x + 1)/(x2 – 1)
(dy/dx) + 2(x + 2)y/(x– 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)/(x– 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)] = (x– 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
= e2log|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 = xex
Solution:
We have,
(dy/dx) – y = xex  ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -1, Q = xex
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)(xex)dx + c
ye-x = ∫xdx + c
ye-x = (x2/2) + c
y = [(x2/2) + c].ex 
This is the required solution.
Question 34.  (dy/dx) + 2y = xe4x
Solution:
We have,
 (dy/dx) + 2y = xe4x   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 2, Q = xe4x
So,
I.F = e∫Pdx
= e2∫dx
= e2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = ∫(e2x).(xe4x)dx + c
y(e2x) = ∫xe6xdx + c
y(e2x) = x∫e6xdx – ∫{(dx/dx)∫e6xdx}dx + c
e2xy = (xe6x)/6 – ∫(e6x/6)dx + c
e2xy = (xe6x)/6 – e6x/36 + c
y = (xe4x)/6 – e4x/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 = 2y+ 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 = emx, m is a given real number
Solution:
We have,
(dy/dx) + 3y = emx      ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 3, Q = emx 
So,
I.F = e∫Pdx
= e∫3dx
= e3x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e3x) = ∫(e3x).(emx)dx + c
y(e3x) = ∫e(3+m)xdx + c
y(e3x) = e(m+3)x/(m + 3) + c
y = emx/(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 = x
(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
= elog|x|
= x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(x) = ∫(x)(x3)dx + c
xy = ∫x+ 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
= elog|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) = x+ 2
Solution:
We have,
(dy/dx) – 2xy/(1 + x2) = x+ 2   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = -2x/(1 + x2), Q = x+ 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)](x+ 2)dx + c
y/(1 + x2) = ∫[(x+ 2)/(x+ 1)]dx + c
\frac{y}{x^2+1}=∫\frac{x^2+1+1}{x^2+1}dx+c
y/(1 + x2) = ∫dx + ∫dx/(x+ 1) + c
y/(x+ 1) = x + tan-1x + c
y = (x + tan-1x + c)(x+ 1)
This is the required solution.
Question 36(vii). Find one-parameter families of solution curves of the following differential equation (dy/dx) + ycosx = esinxcosx
Solution:
We have,
(dy/dx) + ycosx = esinxcosx    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = cosx, Q = esinxcosx
So,
I.F = e∫Pdx
= e∫cosxdx
= esinx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y[(esinx) = ∫(esinx)(esinxcosx)dx + c
Let, sinx = z
ON differentiating both sides we have,
cosxdx = dz
ye= ∫e2zdz + c
ye= (e2z/2) + c
y = ez/2 + ce – z
y = (esinx/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) = cey
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}
= etanx
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(etanx) = ∫(etanx)(tanx.sec2x)dx + c
Let, tanx = z
On differentiating both sides we have,
sec2xdx = dz
y(ez) = ∫zezdz + c
y(ez) = z∫ezdz – ∫{(dz/dz)∫ezdz}dz
y(ez) = ze– ∫ezdz + c
y(ez) = ze– e+ 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
= ey
The solution of a differential equation is,
x(I.F) = ∫Q(I.F)dy + c
x(ey) = ∫e-ysce2yeydy + c
x(ey) = ∫sec2ydy + c
x(ey) =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
= elog|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)+ 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
= e2∫dx/x
= e2logx
= 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)x+ c 
y = (x/16)(4logx – 1) + c/x2
This is the required solution.
Question 37. Solve the following using the initial value problem:-
(i). y’ + y = ex, y(0) = (1/2)
Solution:
We have,
y’ + y = ex
dy/dx + y = ex    ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Py = Q
Where, P = 1, Q = ex
So, I.F = e∫Pdx
= e∫dx
= ex
The solution of differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(ex) = ∫ex.exdx + c
y(ex) = (1/2)e2x + c
At t = 0, y = (1/2)
(1/2)e= (1/2)e+ c
c = 0
y(ex) = (1/2)e2x
y = (ex/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
= e2x
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
y(e2x) = ∫e-2xsinx.(e2x)dx + c
y(e2x) = ∫sinxdx + c
y(e2x) = -cosx + c
At x = 0, y = 0
0 = -1 + c
c = 1
y(e2x) = 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)
= elog(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/(y+ 1)]x = e^{-tan^{-1}y} /(y+ 1)   ………..(i)
The given equation is a linear differential equation of the form
(dy/dx) + Px = Q
Where, P = 1/(y+ 1), Q = e^{-tan^{-1}y} /(y+ 1)
So,
I.F = e∫Pdy
=e^{∫\frac{1}{y^2+1}dy}
= etan-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} /(y+ 1)]e^{-tan^{-1}y} dx + c
x(e^{-tan^{-1}y} ) = ∫dy/(1 + y2) + c
x(etan-1y) = tan-1y + c
At x = 0, y = 0
0*e= 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
= elog|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 = x+ 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
= elog|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
= elog|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) = 2x+ c
At x = π/2, y = 0,
0 = 2(π/2)+ c
c = -π2/2
y(sinx) = 2x– 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
= e2log|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
= elog|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
= elog|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
= e2∫dx/x
= e2logx
= x2
The solution of a differential equation is,
y(I.F) = ∫Q(I.F)dx + c
yx= ∫x2.xdx + c
yx= ∫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) + cex
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
= elog|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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