Exact Equation, integrating factor

**HelloWorld2** · September 27th 2010, 06:12 PM

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

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

**TheEmptySet** · September 27th 2010, 06:53 PM

Originally Posted by HelloWorld2

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

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

So you are looking for a function $\phi$ of one variable either x or y that when you multiply the equation that makes it exact.

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

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

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

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

**Danny** · September 28th 2010, 03:12 PM

I might add that if you write your ODE as

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

it's linear in $x$ .

**Krizalid** · September 29th 2010, 12:56 PM

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

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

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

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

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