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Q.
A function $y=f(x)$ satisfies $(x+1) \cdot f^{\prime}(x)-2\left(x^2+x\right) f(x)=\frac{e^{x^2}}{(x+1)}, \forall x>-1$ If $f(0)=5$, then $f(x)$ is
Differential Equations
Solution:
$ f ^{\prime}( x )-\frac{2 x ( x +1)}{ x +1} f ( x )=\frac{ e ^{ x ^2}}{( x +1)^2} $
$\text { I. F. }= e ^{\int-2 xdx }= e ^{- x ^2} $
$\therefore f ( x ) \cdot e ^{- x ^2}=\int \frac{ dx }{( x +1)^2} \Rightarrow f ( x ) \cdot e ^{- x ^2}=-\frac{1}{ x +1}+ C$
$ \text { at } x =0, f (0)=5 \Rightarrow C =6$
$\therefore f ( x )=\left(\frac{6 x +5}{ x +1}\right) \cdot e ^{ x ^2} $