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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 :

JEE · Math · previous-year question

  1. A.1 $$-$$ e
  2. B.0
  3. C.$${1 \over 2}$$
  4. D.$${4 \over e} - e$$correct

Answer

D. $${4 \over e} - e$$

Explanation

$$x{{dy} \over {dx}} + 2y = x{e^x},\,\,y(1) = 0$$ $${{dy} \over {dx}} + {2 \over x}y = {e^x}$$, then $${e^{\int {{2 \over x}dx} }}dx = {x^2}$$ $$y\,.\,{x^2} = \int {{x^2}{e^x}dx} $$ $$y{x^2} = {x^2}{e^x} - \int {2x{e^x}dx} $$ $$ = {x^2}{e^x} - 2(x{e^x} - {e^x}) + c$$ $$y{x^2} = {x^2}{e^x} - 2x{e^x} + 2{e^x} + c$$ $$y{x^2} = ({x^2} - 2x + 2){e^x} + c$$ $$0 = e + c \Rightarrow c = - e$$ $$y(x)\,.\,{x^2} - {e^x} = {(x - 1)^2}{e^x} - e$$ $$z(x) = {(x - 1)^2}{e^x} - e$$ For local maximum $$z'(x) = 0$$ $$\therefore$$ $$2(x - 1){e^x} + {(x - 1)^2}{e^x} = 0$$ $$\therefore$$ $$x = - 1$$ And local maximum value $$ = z( - 1)$$ $$ = {4 \over e} - e$$

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