1. Exact Equation, integrating factor

Find an integration factor, and use it to solve the following DE:

$\displaystyle ydx+(2xy-e^{-2y})dy=0$

2. Originally Posted by HelloWorld2
Find an integration factor, and use it to solve the following DE:

$\displaystyle ydx+(2xy-e^{-2y})dy=0$
So you are looking for a function $\displaystyle \phi$ of one variable either x or y that when you multiply the equation that makes it exact.

So lets multiply by $\displaystyle \phi(y)$ to get

$\displaystyle y\phi(y)dx+(2xy\phi(y)-e^{-2y}\phi(y))dy=0$

Now we take the partial derivatives to see if it is exact and we get

$\displaystyle \displaystyle \phi +y\phi_y=2y\phi \iff \left(2-\frac{1}{y}\right)dy=\frac{d\phi}{\phi} \implies \phi=\frac{e^{2y}}{y}$

$\displaystyle \dfrac{dx}{dy} + 2x = \dfrac{e^{-2y}}{y}$

it's linear in $\displaystyle x$.

4. check your notes, there're two basic cases when finding an integrating factor:

if $\displaystyle \displaystyle\frac{1}{N}\left( \frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} \right)=h(x),$ then an integrating is $\displaystyle \mu(x)=e^{\int h(x)\,dx}.$

if $\displaystyle \displaystyle\frac{1}{M}\left(\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y} \right)=h(y),$ then an integrating is $\displaystyle \mu(y)=e^{\int h(y)\,dy}.$

by the form of your ODE, i'd choose the second one.