If the solution of the differential equation [dy dx + e^x( x^2 - 2 )y = ( x^2 - 2x )( x^2 …

Question

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 \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.

Accepted Answer

$$\because$$ $${{dy} \over {dx}} + {e^x}({x^2} - 2)y = ({x^2} - 2x)({x^2} - 2){e^{2x}}$$

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 $({x^2} - 2x){e^x} = t$

$$\therefore$$ $({x^2} - 2){e^x}dx = dt$

$$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt} $$

$$y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c$$

$$\therefore$$ $y(0) = 0$

$$\therefore$$ $c = 1$

$$\therefore$$ $$y = ({x^2} - 2x - 1) + {e^{(2x - {x^2}){e^x}}}$$

$$\therefore$$ $y(2) =  - 1 + 1 = 0$

Solutions · 1