# y\ln x-xy'=0?

$y\mathrm{ln}x-x{y}^{\prime }=0$?
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Raymond Foley
$y\mathrm{ln}x-x{y}^{\prime }⇔y\mathrm{ln}x=x\frac{dy}{dx}⇔\frac{dy}{y}=\frac{\mathrm{ln}x}{x}dx$
$\int \frac{dy}{y}=\int \frac{\mathrm{ln}x}{x}dx$
$\mathrm{ln}|y|=\frac{{\mathrm{ln}}^{2}x}{2}+C$
${e}^{\mathrm{ln}|y|}={e}^{\frac{{\mathrm{ln}}^{2}x}{2}+C}$
$|y|={e}^{\frac{{\mathrm{ln}}^{2}x}{2}}×{e}^{C}\wedge {e}^{C}\phantom{\rule{0.222em}{0ex}}={c}_{1}$
$y=c{e}^{\frac{{\mathrm{ln}}^{2}x}{2}}$

###### Not exactly what you’re looking for?
Daniel Cormack
$\int \frac{\mathrm{ln}x}{x}dx$
Let $\mathrm{ln}x=u$
thus, we have
$\frac{dx}{x}=du$
$\int \frac{\mathrm{ln}x}{x}dx=\int udu$
$=\frac{{u}^{2}}{2}+C$
$=\frac{{\mathrm{ln}}^{2}x}{2}+C$
Therefore, $y=c{e}^{\frac{{\mathrm{ln}}^{2}x}{2}}$ is the solution