If $y=y\left(x\right)$ is the solution of the differential equation $x\frac{dy}{dx}+2y=x{e}^x,y\left(1\right)=0$ then the local maximum value of the function $z\left(x\right)=x^{2}y\left(x\right)-e^x,x\in R$ is

Given,

$\frac{dy}{dx}+\frac{2y}{x}=e^x$

$I.F.=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2$

$\therefore$ General solution: $y\left(x^2\right)=\int e^x\left(x^2\right)dx$

We know that

$\int e^{x}f\left(x\right)dx==e^x\left[f\left(x\right)-f^{\text{'}}\left(x\right)+f^{\text{'}\text{'}}\left(x\right)-f^{\text{'}\text{'}\text{'}}\left(x\right)\left.+\cdots +{\left(-1\right)}^n{f}^n\left(x\right)\right)\right]+C$

So, $y\left(x^2\right)=e^x\left[x^2-2x+2\right]+C ...\left(\mathrm{i}\right)$

Given $y\left(1\right)=0\Rightarrow \left(0\right)\left(1\right)=e^1\left[1-2\left(1\right)+2\right]+C$

$\Rightarrow C=-e$

$\therefore y=\frac{e^x}{x^2}\left[x^2-2x+2\right]-e$     (from eq $\left(\mathrm{i}\right)$)

Hence, $z\left(x\right)=x^2\left(\frac{e^x}{x^2}\right)\left[x^2-2x+2\right]-e-e^x$

$=e^x\left[x^2-2x+2\right]-e-e^x$ 

$z\left(x\right)=e^x\left[x^2-2x+1\right]-e$

$z^{\text{'}}\left(x\right)=e^x\left(2x-2\right)+\left(x^2-2x+1\right)e^x$

$z^{\text{'}}\left(x\right)=e^x\left[x^2-1\right]$
To find local maxima, put $z^{\text{'}}\left(x\right)=0$

$\Rightarrow e^x\left(x^2-1\right)=0$ $\Rightarrow \left(x-1\right)\left(x+1\right)=0$  $\left(\because e^x>0, \forall x\in R\right)$

$\Rightarrow x=1,-1$

It is clear from the sign scheme method, $z\left(x\right)$ has local maximum value at $x=-1$ and has local minimum value at $x=1$.

$\therefore$  The local maximum value is

$z\left(-1\right)=e^{-1}\left[{\left(-1\right)}^2-2\left(-1\right)+1\right]-e$

$=\frac{1}{e}\left[1+2+1\right]-e$ $=\frac{4}{e}-e$