If y = y(x) is the solution of the differential equation [xdy dx + 2y = x,e^x], y(1) = 0 t…

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 :

Accepted Answer

$$x{{dy} \over {dx}} + 2y = x{e^x},\,\,y(1) = 0$$

$${{dy} \over {dx}} + {2 \over x}y = {e^x}$$, 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 $= z( - 1)$

$$ = {4 \over e} - e$$

Solutions · 1