 # Solve differential equationdy/dx= x/(ye^(x+y^2)) Jaden Easton 2020-12-28 Answered

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

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$dy/dx=x/\left(y{e}^{x+{y}^{2}}\right)$
$=>dy/dx=x/\left(y{e}^{x}\ast {e}^{{y}^{2}}\right)$
$=>y{e}^{\left(}{y}^{2}\right)dy=x/{e}^{x}dx$
$y{e}^{{y}^{2}}dy=x{e}^{-x}dx$
$\int {e}^{{y}^{2}}ydy=\int x{e}^{-x}dx+c$
where C is integrating constant
$\int {e}^{{y}^{2}}ydy=\int x{e}^{-x}dx+c$
${y}^{2}=t$
$2ydy=dt$
$ydy=\left(dt\right)/2$
$⇒\frac{1}{2}\int {e}^{t}dt=\int x{e}^{-x}dx+c$(integrating by part)
$⇒\frac{1}{2}{e}^{t}=-x{e}^{-x}+\int {e}^{-x}dx+c$
$1/2{e}^{{y}^{2}}=-x{e}^{-x}+\left(-{e}^{-x}+c\therefore \left(t={y}^{2}\right)\right)$
$1/2{y}^{2}=-{e}^{-}x\left[x+1\right]+c$
$1/2{y}^{2}+{e}^{-}n\left(x+1\right)=c$
$1/2{y}^{2}=-{e}^{-}x\left(x+1\right)+c$
${y}^{2}/2+{e}^{-}x\left(x+1\right)=C$

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