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

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

$\displaystyle (xy'-1)\ln x=2y$
$\displaystyle y'x\ln x -\ln x=2y$
$\displaystyle y'x\ln x -2y=\ln x$
$\displaystyle y'-\frac{2y}{x\ln x}=\frac{1}{x}$

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

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

2. ## Re: First-order linear DE--help w/ integrating factor

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

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

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 \displaystyle \begin{align*} e^{\int{-\frac{2}{x\ln{x}}\,dx}} \end{align*}. You can evaluate this integral using the substitution \displaystyle \displaystyle \begin{align*} u = \ln{x} \implies du = \frac{1}{x}\,dx \end{align*}.

3. ## Re: First-order linear DE--help w/ integrating factor

Ah, that did it. Thanks.