# Solve the differential equationx dy/dx-y= x^2lnx

Solve the differential equation $xdy/dx-y={x}^{2}\mathrm{ln}x$

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Alix Ortiz

Divide both sides by x and further simplify it
$x/xdy/dx-y/x=\left({x}^{2}\mathrm{ln}x\right)/x$
$dy/dx-y/x=x\mathrm{ln}x$
It is of the form $dy/dx+P\left(x\right)y=Q\left(x\right)$
Now find the integrating factor by using $\left(I.F.\right)={e}^{\int P\left(x\right)dx}$
Hence, $I.F.={e}^{-\int 1/xdx}$
$={e}^{-\mathrm{ln}x}$
$={e}^{\mathrm{ln}1/x}$
$=1/x$
Hence, $I.F.=1/x$
$y\ast \left(I.F.\right)=\int \left(I.F.\right)\ast Q\left(x\right)dx+c$
Hence, $y\ast 1/x=\int 1/x\ast x\mathrm{ln}xdx+c$
$y\ast 1/x=\int \mathrm{ln}xdx+c$
$y\ast 1/x=\int \mathrm{ln}x\ast 1dx+c$
$y\ast 1/x=\mathrm{ln}x+x+c$
Simplify
$y\ast 1/x=x\left(\mathrm{ln}x+1\right)+c$
$y={x}^{2}\left(\mathrm{ln}x+1\right)+cx$