Originally Posted by

**sterps** I need some help with this question.

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

The first part i have done.

$\displaystyle (1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0$

Use $\displaystyle Mdx+Ndy=0$

$\displaystyle dM/dy = (1+2xe^{-y})dx$

$\displaystyle 0+(0.e^{-y})+(2xe^{-y})$

$\displaystyle 2xe^{-y}$

$\displaystyle dN/dx = (1+x^2)e^{-y}dy$

$\displaystyle 0+(2x.e^{-y})+(x^2.0)$

$\displaystyle 2xe^{-y}$

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!