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Q.
The solution of the differential equation, $e ^{ x }( x +1) dx +\left( ye ^{ y }- xe ^{ x }\right) dy =0$ with initial condition $f(0)=0$, is
Differential Equations
Solution:
$\text { put } x e^x=t $
$\left(e^x+x e^x\right) \frac{d x}{d y}=\frac{d t}{d y}$
$\therefore \frac{ dt }{ dy }+\left( ye ^{ y }- t \right)=0 \Rightarrow \frac{ dt }{ dy }- t + ye ^{ y }=0$
I.F. $ e ^{-\int d y}= e ^{- y }$
$t \cdot e ^{- y }=-\int ye ^{ y } e ^{- y } dy $
$x e ^{ x } e ^{- y }=-\frac{ y ^2}{2}+ C$
$f(0)=0 \Rightarrow C =0 ; 2 xe ^{ x } e ^{- y }+ y ^2=0$