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Q.
If $y = y ( x )$ is the solution of the differential equation $x \frac{ dy }{ dx }+2 y = xe ^{ x }, y (1)=0$ then the local maximum value of the function $z(x)=x^{2} y(x)-e^{x}, x \in R$ is :
$x \frac{d y}{d x}+2 y=x e^{x}$
$\frac{d y}{d x}+\frac{2 y}{x}=e^{x}$
$I.F. =x^{2}$
$ y .x x^{2}=\int x^{2} e^{x} d x$
$=\int e^{x}\left(x^{2}+2 x-2 x-2+2\right) d x$
$y x^{2} =e^{x}\left(x^{2}-2 x+2\right)+c$
$y(1) =0$
$0=e(1+0)+c$
$c =-e$
$z(x) =x^{2} y(x)-e^{x}$
$=e^{x}\left(x^{2}-2 x+2\right)-e-e^{x}$
$=e^{x}(x-1)^{2}-e$
$\frac{ dz }{ dx }= e ^{ x } \cdot 2( x -1)+ e ^{ x }( x -1)^{2}=0$
$x ^{ x }( x -1)(2+ x -1)=0$
$e ^{ x }( x -1)( x +1)=0$
$x =-1,1$
$x =-1$ local maxima. Then maximum value is
$z (-1)=\frac{4}{ e }- e$
