# First-order linear DE--help w/ integrating factor

• Aug 21st 2012, 08:44 PM
phys251
First-order linear DE--help w/ integrating factor
$(xy'-1)\ln x=2y$
$y'x\ln x -\ln x=2y$
$y'x\ln x -2y=\ln x$
$y'-\frac{2y}{x\ln x}=\frac{1}{x}$

so $P(x) = e^{\int x\ln x~dx}$

Help with that integration, please? Should I try substitution, or should I go with integration by parts?
• Aug 21st 2012, 09:06 PM
Prove It
Re: First-order linear DE--help w/ integrating factor
Quote:

Originally Posted by phys251
$(xy'-1)\ln x=2y$
$y'x\ln x -\ln x=2y$
$y'x\ln x -2y=\ln x$
$y'-\frac{2y}{x\ln x}=\frac{1}{x}$

so $P(x) = e^{\int x\ln x~dx}$

Help with that integration, please? Should I try substitution, or should I go with integration by parts?

First of all, your integrating factor is actually \displaystyle \begin{align*} e^{\int{-\frac{2}{x\ln{x}}\,dx}} \end{align*}. You can evaluate this integral using the substitution \displaystyle \begin{align*} u = \ln{x} \implies du = \frac{1}{x}\,dx \end{align*}.
• Aug 24th 2012, 01:48 PM
phys251
Re: First-order linear DE--help w/ integrating factor
Ah, that did it. Thanks.