1. [SOLVED] Initial value problem

I need some help with this question.

Show that the equation $(1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$ is exact and hence find the solution of the initial value problem $dy/dx=\frac{2x+e^y}{1+x^2}$ on the interval 0<_x<_1

The first part i have done.
$(1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$
Use $Mdx+Ndy=0$

$dM/dy = (1+2xe^{-y})dx$
$0+(0.e^{-y})+(2xe^{-y})$
$2xe^{-y}$

$dN/dx = (1+x^2)e^{-y}dy$
$0+(2x.e^{-y})+(x^2.0)$
$2xe^{-y}$
therefore the eqn. is exact.

Its just the next part that i need some help with, a nudge in the right direction would be great!

2. Originally Posted by sterps
I need some help with this question.

Show that the equation $(1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$ is exact and hence find the solution of the initial value problem $dy/dx=\frac{2x+e^y}{1+x^2}$ on the interval 0<_x<_1

The first part i have done.
$(1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$
Use $Mdx+Ndy=0$

$dM/dy = (1+2xe^{-y})dx$
$0+(0.e^{-y})+(2xe^{-y})$
$2xe^{-y}$

$dN/dx = (1+x^2)e^{-y}dy$
$0+(2x.e^{-y})+(x^2.0)$
$2xe^{-y}$
therefore the eqn. is exact.

Its just the next part that i need some help with, a nudge in the right direction would be great!
The solution is $M(x, y) = C$ where

$\frac{\partial M}{\partial x} = 1 + 2x e^{-y}$ .... (1)

$\frac{\partial M}{\partial y} = -(1 + x^2) e^{-y} = -e^{-y} - x^2 e^{-y}$ .... (2)

Solve equations (1) and (2) simultaneously for M.