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If the solution of the differential equation $${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$$ satisfies $$y(0) = 0$$, then the value of y(2) is _______________.

JEE · Math · previous-year question

  1. A.$$-$$1
  2. B.1
  3. C.0correct
  4. D.e

Answer

C. 0

Explanation

$$\because$$ $${{dy} \over {dx}} + {e^x}({x^2} - 2)y = ({x^2} - 2x)({x^2} - 2){e^{2x}}$$ Here, $$I.F. = {e^{\int {{e^x}({x^2} - 2)dx} }}$$ $$ = {e^{({x^2} - 2x){e^x}}}$$ $$\therefore$$ Solution of the differential equation is $$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {({x^2} - 2x)({x^2} - 2){e^{2x}}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$ $$ = \int {({x^2} - 2x){e^x}\,.\,({x^2} - 2){e^x}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$ Let $$({x^2} - 2x){e^x} = t$$ $$\therefore$$ $$({x^2} - 2){e^x}dx = dt$$ $$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt} $$ $$y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c$$ $$\therefore$$ $$y(0) = 0$$ $$\therefore$$ $$c = 1$$ $$\therefore$$ $$y = ({x^2} - 2x - 1) + {e^{(2x - {x^2}){e^x}}}$$ $$\therefore$$ $$y(2) = - 1 + 1 = 0$$

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