JEE Math Practice 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 :
- A.1 $$-$$ e
- B.0
- C.$${1 \over 2}$$
- D.$${4 \over e} - e$$
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