"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 :"

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Accepted Answer

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

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

$= {4 \over e} - e$

"

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 :

Solution

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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 :

Solution

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More Differential Equations questions

Drill 25 more like these. Every day. Free.

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 :