# Math Help - Another general solution

1. ## Another general solution

find the general solution of the first order diff. equation.

e^y + 2xy + (xe^y + x^2)dy/dy = 0

I divided accross by (xe^y + x^2) to get 2ln(y) - ln(x) + dy/dx = 0
Not sure where to take it from here.Help would be greatly appreciated.

2. Originally Posted by matty888
find the general solution of the first order diff. equation.

e^y + 2xy + (xe^y + x^2)dy/dy = 0

I divided accross by (xe^y + x^2) to get 2ln(y) - ln(x) + dy/dx = 0
Not sure where to take it from here.Help would be greatly appreciated.
I think what you meant is

$
e^y + 2xy + (xe^y + x^2)\frac{dy}{dx} = 0
$

Really not sure what you did but multiplying by dx gives

$
(e^y + 2xy)dx + (xe^y + x^2)dy = 0
$

and this ODE is exact meaning that there exist a $\phi$ such that

$
d \phi = \frac{\partial \phi}{\partial x}\; dx + \frac{\partial \phi}{\partial y}\; dy = (e^y + 2xy)dx + (xe^y + x^2)dy
$