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

Question

If $y = y(x)$ is the solution of the differential equation

$x{{dy} \over {dx}} + 2y = x\,{e^x}$, $y(1) = 0$ then the local maximum value

of the function $z(x) = {x^2}y(x) - {e^x},\,x \in R$ is :

1 $-$ e

${1 \over 2}$

${4 \over e} - e$

Solution

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