 # Solve differential equation(x^2lnx)y'+xy=0 snowlovelydayM 2020-11-26 Answered

Solve differential equation $\left({x}^{2}\mathrm{ln}x\right){y}^{\prime }+xy=0$

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Use the variable separation method to solve the differential equation
$\int 1/xdx=\mathrm{ln}\left(x\right)+c$
$\left({x}^{2}\mathrm{ln}x\right){y}^{\prime }+xy=0$
$\left({x}^{2}\mathrm{ln}x\right)dy/dx=-xy$
$dy/y=-x/\left({x}^{2}\mathrm{ln}x\right)dx$
$dy/y=-1/\left(x\mathrm{ln}x\right)dx$
integrated both sides
$\mathrm{ln}\left(y\right)=-\mathrm{ln}\left(\mathrm{ln}x\right)+\mathrm{ln}\left(c\right)$
$\mathrm{ln}\left(y\right)=\mathrm{ln}\left(\mathrm{ln}x{\right)}^{-1}+\mathrm{ln}\left(c\right)=\mathrm{ln}\left(1/\left(\mathrm{ln}\left(x\right)\right)\right)+\mathrm{ln}\left(c\right)$ ${:}^{\prime }\mathrm{ln}a+\mathrm{ln}b=\mathrm{ln}ab$

$y=c/\left(\mathrm{ln}\left(x\right)\right)$

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