Answer to Question #231549 in Differential Equations for Phyroehan

Question

y ln x ln y dx + dy =0

Answer 1

ylnxlnydx=-dy

lnxdx=-dy/(ylny)

$\intop lnxdx=fg-\int f'g$

f=lnx

g'=1

f'=1/x

g=x

$\int lnxdx=xlnx- \int 1dx=x(lnx-1)+C$

$\int \frac{dy}{ylny}$

u=lny

du/dy=1/y

dy=ydu$\int \frac{1}{u} du= lnu$

$\int \frac{dy}{ylny}=ln(lny)+C$

$x(lnx-1)+C=ln(lny)$

$y=e^{e^{x(lnx-1)+C}}$