JEE MAIN - Mathematics (2022 - 26th June Evening Shift - No. 8)
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 :
$$x{{dy} \over {dx}} + 2y = x\,{e^x}$$, $$y(1) = 0$$ then the local maximum value
of the function $$z(x) = {x^2}y(x) - {e^x},\,x \in R$$ is :
1 $$-$$ e
0
$${1 \over 2}$$
$${4 \over e} - e$$
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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