Solution
given the differential equation:
(x2+y2+2x)dx+2ydy=0 ...(1)
let f(x,y)=x2+y2+2x
⟹g(y)=2y
so
f'y(x,y)=2y
g′(x)=0
using Integration factor method,
dxdu=g(fy−gx)u
dxdu=2y(2y−0)u
dxdu=2y(2y)u
dxdu=u
on solving we get integrated factor
u=ex
Multiply equation 1 by integrated factor
ex(x2+y2+2x)dx+2yexdy=0
let M=(x2+y2+2x)ex
N=2yex
dydM=2yex
dxdN=2yex
Hence dydM=dxdN
Solution of above equation is given by,
Let I(x,y) be a implict function
I(x,y)=∫M(x,y)dx
=∫ex(x2+y2+2x)dx
=x2ex−2xex+2ex+y2ex+2xex−2ex
=x2ex+y2ex +f(y)
Let differentiate I(x,y) with respect to y
dydI= N(x,y)
2yex+dydf=2yex
dydf=0=constant
I(x,y)=x2ex+y2ex+c
at x=y=1
0=e+e+cc+2e=0c=−2e
hence this is the required solution is;
x2ex+y2ex−2e=0
x2+y2=2e1−x .
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