# Find general solution of the following differential equation dy/dx=(2y^2+x^2e^(-(y/x)^2))/(2xy)

Find general solution of the following differential equation $dy/dx=\left(2{y}^{2}+{x}^{2}{e}^{\left(}-\left(y/x{\right)}^{2}\right)\right)/\left(2xy\right)$
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dieseisB

$y=vx=>y/x=v$
$dy/dx=v+x\left(dv\right)/dx$
$v+x\left(dv\right)/dx=\left(2\left(vx{\right)}^{2}+{x}^{2}{e}^{\left(}-\left(v{\right)}^{2}\right)\right)/\left(2x\left(vx\right)\right)$
$v+x\left(dv\right)/dx=\left(2{v}^{2}{x}^{2}+{x}^{2}{e}^{\left(}-\left(v{\right)}^{2}\right)\right)/\left(2v{x}^{2}\right)$
$v+x\left(dv\right)/dx=\left({x}^{2}\left(2{v}^{2}+{e}^{\left(}-\left(v{\right)}^{2}\right)\right)\right)/\left(2v{x}^{2}\right)$ (take ${x}^{2}$ as common factor from numerator)
$v+x\left(dv\right)/dx=\left(2{v}^{2}+{e}^{\left(}-\left(v{\right)}^{2}\right)\right)/\left(2v\right)$ (divide ${x}^{2}/{x}^{2}=1$)
Substract v from both sides and further simplify it
$v+x\left(dv\right)/dx-v=\left(2{v}^{2}+{e}^{\left(}-\left(v{\right)}^{2}\right)\right)/\left(2v\right)-v$
$x\left(dv\right)/dx=\left(2{v}^{2}{e}^{\left(}-\left(v{\right)}^{2}\right)-2{v}^{2}\right)/\left(2v\right)$
$x\left(dv\right)/dx={e}^{\left(}-\left(v{\right)}^{2}\right)/\left(2x\right)$
$\left(2v\right)/{e}^{\left(}-\left(v{\right)}^{2}\right)dv=dx/x$
Integrate both sides
$\int 2v{e}^{{v}^{2}}dv=\int dx/x+c$
Substitute ${v}^{2}=t$
$=>2vdv=dt$
Hence, $\int {e}^{t}dt=\int dx/x+c$
${e}^{t}=\mathrm{ln}x+c$
$t=\mathrm{ln}\left(\mathrm{ln}x+c\right)$ (use property ${e}^{x}=y=>x=\mathrm{ln}y$)
Substitute back $t={v}^{2}$
${v}^{2}=\mathrm{ln}\left(\mathrm{ln}x+c\right)$
Substitute back $v=y/x$
$\left(y/x{\right)}^{2}=\mathrm{ln}\left(\mathrm{ln}x+c\right)$
${y}^{2}/x=\mathrm{ln}\left(\mathrm{ln}x+c\right)$
${y}^{2}={x}^{2}\mathrm{ln}\left(\mathrm{ln}x+c\right)$
$y=+-\sqrt{{x}^{2}\mathrm{ln}\left(\mathrm{ln}x+c\right)}$
$y=±x\sqrt{\mathrm{ln}\left(\mathrm{ln}x+C\right)}$

Jeffrey Jordon