Integrating factors

• Sep 11th 2010, 08:13 PM
A Beautiful Mind
Integrating factors
So, some guidance would be helpful.

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

$\frac{1-2y}{-y}$

$e^{^-{\int \frac{1}{y}-2}}$

$\frac{e^{2y}}{y}$ is the integrating factor.

And upon solving, the answer is: $C = xe^{2y}-ln|y|$?
• Sep 11th 2010, 08:40 PM
Prove It
$y\,dx + (2xy - e^{-2y})\,dy = 0$

$y\,\frac{dx}{dy} + 2xy - e^{-2y} = 0$

$\frac{dx}{dy} + 2x - \frac{e^{-2y}}{y} = 0$

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

The integrating factor is $e^{\int{2\,dy}} = e^{2y}$

So multiplying through by the integrating factor gives

$e^{2y}\,\frac{dx}{dy} + 2e^{2y}x = \frac{1}{y}$

$\frac{d}{dy}(e^{2y}x) = \frac{1}{y}$

$e^{2y}x = \int{\frac{1}{y}\,dy}$

$e^{2y}x = \ln{|y|} + C$

$x = e^{-2y}\ln{|y|} + Ce^{-2y}$.
• Sep 11th 2010, 09:02 PM
A Beautiful Mind
If I had solved for x at the end, I would've gotten the same answer. Is my integrating factor incorrect?
• Sep 11th 2010, 09:04 PM
Prove It
Quote:

Originally Posted by A Beautiful Mind
If I had solved for x at the end, I would've gotten the same answer. Is my integrating factor incorrect?

I don't see how you got your Integrating Factor to be honest...

$\frac{My-Nx}{-M}$