# Solve differential equation (2y-e^x)dx+xdy=0, x>0

Solve differential equation $\left(2y-{e}^{x}\right)dx+xdy=0$, x>0
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Nathanael Webber

$\left(2y-{e}^{x}\right)dx+xdy=0$
$2y-{e}^{x}+xdy/dx=0$
$\left(2y\right)/x-{e}^{x}/x+dy/dx=0$
$dy/dx+\left(2/x\right)y={e}^{x}/x$
Above differential equation is form of $dy/dx+Py=Q$
Integrating factor will be
$I.F.={e}^{\int Pdx}$
$={e}^{\int 2/xdx}$
$={e}^{2\mathrm{ln}x}$ $={e}^{\mathrm{ln}{x}^{2}}$
$={x}^{2}$
Solution will be given as
$I.F\ast y=\int I.F\ast Qdx$
${x}^{2}y=\int {x}^{2}\ast {e}^{2}/xdx$
${x}^{2}y=\int x{e}^{x}dx$
${x}^{2}y=\left[x\int {e}^{x}dx-\int \left[d/dx\left(x\right)\int {e}^{x}dx\right]\right]dx$
${x}^{2}y=\left[x{e}^{x}-{e}^{x}\right]+c$
$y=1/{x}^{2}\left(x{e}^{x}-{e}^{x}+c\right)$