Elleanor Mckenzie

2020-11-07

Solve differential equation $xydy-{y}^{2}dx=\left(x+y{\right)}^{2}{e}^{\left(}-y/x\right)$

jlo2niT

$xydy-{y}^{2}dx=\left(x+y{\right)}^{2}{e}^{\left(}-y/x\right)$
$xydy/dx-{y}^{2}=\left(x+y{\right)}^{2}{e}^{\left(}-y/x\right)$ (1)
Put $y/x=t=>dy/dx=y+x\left(dt\right)/dx$
Now the equation (1) becomes
$x\ast xt\left[t+x\left(dt\right)/dx\right]-{x}^{2}{t}^{2}={x}^{2}\left(1+t{\right)}^{2}{e}^{-}t$
$t\left[t+x\left(dt\right)/dx\right]-{t}^{2}=\left(1+t{\right)}^{2}{e}^{t}$
$xt\left(dt\right)/dx=\left(1+t{\right)}^{2}{e}^{t}$
$\left({e}^{t}tdt\right)/\left(1+t{\right)}^{2}=dx/x$
$\left({e}^{t}\left(\left(t+1\right)-1\right)dt\right)/\left(1+t{\right)}^{2}=dx/x$
${e}^{t}\left(1/\left(1+t\right)-1/\left(1+t{\right)}^{2}\right)dt=dx/x$ (2)
Integrating on the both sides
$\int {e}^{t}\left(1/\left(1+t\right)-1/\left(1+t{\right)}^{2}\right)dt=\int dx/x$
Use the standard formula of integration
$\int {e}^{x}\left(f\left(x\right)+{f}^{\prime }\left(x\right)\right)dx={e}^{x}f\left(x\right)+C$
${e}^{t}/\left(\left(1+t\right)\right)=\mathrm{ln}\left(x\right)+\mathrm{log}\left(C\right)$
${e}^{y/x}/\left(1+y/x\right)=\mathrm{ln}\left(Cx\right)$
${e}^{y/x}=\left(1+y/x\right)\mathrm{ln}\left(Cx\right)$

Jeffrey Jordon