# Solve differential equation xy'-2x^2y= e^(x^2)

Solve differential equation$x{y}^{\prime }-2{x}^{2}y={e}^{\left(}{x}^{2}\right)$
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Yusuf Keller

Rewrite the above differential equation dividing throughout by x as follows
$\left(x{y}^{\prime }\right)/x-\left(2{x}^{2}y\right)/x={e}^{\left(}{x}^{2}\right)/x$
$=>{y}^{\prime }-2xy={e}^{\left(}{x}^{2}\right)/x$
${y}^{\prime }+p\left(x\right)y=q\left(x\right)withp\left(x\right)=-2x$, $q\left(x\right)={e}^{\left(}{x}^{2}\right)/x$
${y}^{\prime }+p\left(x\right)y=q\left(x\right)$
$y\ast {e}^{\left(\int p\left(x\right)dx\right)}=\int \left({e}^{\left(\int p\left(x\right)dx\right)}\ast q\left(x\right)dx\right)+C$
Hence, the solution $-2xy={e}^{\left(}{x}^{2}\right)/x$
$y\ast {e}^{\left(\int p\left(x\right)dx\right)}=\int \left({e}^{\left(\int p\left(x\right)dx\right)}\ast q\left(x\right)dx+C$
$=>y\ast {x}^{\left(int-2xdx\right)}=\int \left({e}^{\left(\int -2xdx\right)\right)}\ast \left({e}^{\left({x}^{2}\right)}/x\right)dx+C$
$=>y\ast {x}^{\left(-{x}^{2}\right)}=\int \left({e}^{\left(-{x}^{2}\right)}\ast \left({e}^{\left({x}^{2}\right)}/x\right)dx+C$
$=>y\ast {x}^{\left(-{x}^{2}\right)}=\int \left(1/x\right)dx+C$
$=>y\ast {x}^{\left(}-{x}^{2}\right)=ln\left(x\right)+C$
$=>y=\left(ln\left(x\right)+C\right)/\left(e-{x}^{2}\right)$
$=>y={e}^{\left(}{x}^{2}\right)\left(ln\left(x\right)+C\right)$
$=>y={e}^{\left(}{x}^{2}\right)ln\left(x\right)+C{e}^{\left(}{x}^{2}\right)$
$x{y}^{\prime }-2{x}^{2}y={e}^{\left(}{x}^{2}\right)$ $y={e}^{\left(}{x}^{2}\right)ln\left(x\right)+C{e}^{\left(}{x}^{2}\right)$

Jeffrey Jordon