Question

solve
(x*e^xy+2y)dy+(y*e^xy+2y)dx=0​

Answer 1

Answer:

The differential equation is

ye

y

x

dx=(xe

y

x

+y

2

)dy

let's take p as p=e

y

x

so differential equation will become

ypdx=(xp+y

2

)dy

differentiating p with respect to x

dp=e

y

x

(

y

dx

−x

y

2

dy

)

⟹y

2

dx

dp

=p(y−x

dx

dy

)

⟹pydx=y

2

dp+pxdy

apply this on above differential equation

xpdy+y

2

dp=xpdy+y

2

dy

⟹dy=dp,

⟹y=p+C,

hence the solution is :

y=e

y

x

+C