If y=y(x) is the solution of the differential | vedclass

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(D) Given the differential equation: $x \frac{d y}{d x}+2 y=x e^{x}$.Dividing by $x$,we get: $\frac{d y}{d x}+\frac{2}{x} y=e^{x}$.This is a linear di

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$\frac{4}{ e }- e$

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(D) Given the differential equation: $x \frac{d y}{d x}+2 y=x e^{x}$.Dividing by $x$,we get: $\frac{d y}{d x}+\frac{2}{x} y=e^{x}$.This is a linear differential equation of the form $\frac{d y}{d x}+P(x)y=Q(x)$,where $P(x)=\frac{2}{x}$ and $Q(x)=e^{x}$.The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln |x|} = x^{2}$.The solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y \cdot x^{2} = \int e^{x} \cdot x^{2} dx + C$.Using integration by parts: $\int x^{2} e^{x} dx = x^{2} e^{x} - \int 2x e^{x} dx = x^{2} e^{x} - 2(x e^{x} - e^{x}) + C = e^{x}(x^{2}-2x+2) + C$.So,$y x^{2} = e^{x}(x^{2}-2x+2) + C$.Given $y(1)=0$,we have $0 = e^{1}(1-2+2) + C$,which gives $C = -e$.Thus,$y x^{2} = e^{x}(x^{2}-2x+2) - e$.Now,$z(x) = x^{2} y(x) - e^{x} = e^{x}(x^{2}-2x+2) - e - e^{x} = e^{x}(x^{2}-2x+1) - e = e^{x}(x-1)^{2} - e$.To find the local maximum,we find the critical points by setting $z'(x) = 0$.$z'(x) = e^{x}(x-1)^{2} + e^{x} \cdot 2(x-1) = e^{x}(x-1)(x-1+2) = e^{x}(x-1)(x+1) = 0$.The critical points are $x=1$ and $x=-1$.Using the first derivative test,$z'(x) > 0$ for $x  0$ for $x > 1$.Thus,$x=-1$ is a point of local maximum.The local maximum value is $z(-1) = e^{-1}(-1-1)^{2} - e = e^{-1}(4) - e = \frac{4}{e} - e$.

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

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