If y=y(x) 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 $z\left(x\right)=x^{2}y\left(x\right)-e^x, x\in R is :$

$$\begin{gathered} x\frac{dy}{dx}+2y=x{e}^x \\ \frac{dy}{dx}+\frac{2y}{x}=e^x \\ It is a linear differential equation \\ I.F. =x^2 \\ y·x^2=\int x^2{e}^{x}dx \\ y{x}^2=e^x\left(x^2-2x+2\right)+c \\ y\left(1\right)=0 \\ 0= e (1+0)+ c \\ c=-e \\ z\left(x\right)=x^{2}y\left(x\right)-e^x \\ =e^x\left(x^2-2x+2\right)-e-e^x \\ =e^x(x-1{)}^2-e \\ \frac{dz}{dx}=e^x·2(x-1)+e^x(x-1{)}^2=0 \\ {x}^x(x-1)(2+x-1)=0 \\ {e}^x(x-1)(x+1)=0 \\ x=-1,1 \end{gathered}$$

$$\begin{gathered} x =-1 local maxima. Then maximum value is \\ z(-1)=\frac{4}{e}-e \end{gathered}$$