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

**phys251** · August 21st 2012, 07:44 PM

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

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?

**Prove It** · August 21st 2012, 08:06 PM

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 $\displaystyle \begin{align*} u = \ln{x} \implies du = \frac{1}{x}\,dx \end{align*}$ .

**phys251** · August 24th 2012, 12:48 PM

Ah, that did it. Thanks.

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