If y = y(x) is the solution of the differential equation x(dy)/(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 ∈ R is

Correct answer: D. (4)/(e) - e Given, (dy)/(dx) + (2y)/(x) = e ^ x I . F . = e ^ ∫ (2)/(x)dx = e ^ 2 x = x ^ 2 ∴ General solution: y( x ^ 2 ) = ∫ e ^ x ( x ^ 2 )dx We know that ∫ e ^ x f(x)dx = = e ^ x [f(x) - f ^ ' ( x ) + f ^ '' (x) - f ^ ''' (x)+ ⋯ + (- 1) ^ n f ^ n (x))] + C So, y( x ^ 2 ) = e ^ x [ x ^ 2 - 2x + 2] + C . . . (i) Given y(1) = 0 ⇒ (0)(1) = e ^ 1 [1 - 2(1) + 2] + C ⇒ C = - e ∴ y = e ^ x x ^ 2 [ x ^ 2 - 2x + 2] - e (from eq (i) ) Hence, z ( x ) = x ^ 2 ( e ^ x x ^ 2 )[ x ^ 2 - 2x + 2] - e - e ^ x = e ^ x [ x ^ 2 - 2x + 2] - e - e ^ x z(x) = e ^ x [ x ^ 2 - 2x + 1] - e z ^ ' (x) = e ^ x (2x - 2) + ( x ^ 2 - 2x + 1) e ^ x z ^ ' (x) = e ^ x [ x ^ 2 - 1] To find local maxima, put z ^ ' (x) = 0 ⇒ e ^ x ( x ^ 2 - 1) = 0 ⇒ (x - 1)(x + 1) = 0 (∵ e ^ x > 0, x ∈ R) ⇒ x = 1, - 1 It is clear from the sign scheme method, z(x) has local maximum value at x = - 1 and has local minimum value at x = 1 . ∴ The local maximum value is z(- 1) = e ^ - 1 [ (- 1) ^ 2 - 2(- 1) + 1] - e = (1)/(e)[1 + 2 + 1] - e = (4)/(e) - e

JEE Main 202226 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 26 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2022 · 26 June · Evening Shift · Mathematics · Differential Equations

If y = y(x) is the solution of the differential equation x(dy)/(dx)...

y = y\left(x\right) is the solution of the differential equation

x\dfrac{dy}{dx} + 2y = x{e}^{x}, y\left(1\right) = 0 then the local maximum value of the function

z\left(x\right) = {x}^{2}y ( x ) - {e}^{x}, x \in R

If y = y(x) is the solution of the differential equation x(dy)/(dx) + 2y = x e ^ x , y(1) = 0 then the local maximum

Options

A $1 - e$
B $0$
C $\dfrac{1}{2}$
D $\dfrac{4}{e} - e$

Correct answer

D. $\dfrac{4}{e} - e$

Step-by-step solution

Given,

$\dfrac{dy}{dx} + \dfrac{2y}{x} = {e}^{x}$

$I . F . = {e}^{∫ \dfrac{2}{x}dx} = {e}^{2\ln x} = {x}^{2}$

$∴$ General solution: $y\left({x}^{2}\right) = ∫ {e}^{x}\left({x}^{2}\right)dx$

We know that

$∫ {e}^{x}f\left(x\right)dx = = {e}^{x}\left[f\left(x\right) - {f}^{'} ( x ) + {f}^{''}\left(x\right) - {f}^{'''}\left(x\right)\left+ ⋯ + {\left(- 1\right)}^{n}{f}^{n}\left(x\right)\right)\right] + C$

So, $y\left({x}^{2}\right) = {e}^{x}\left[{x}^{2} - 2x + 2\right] + C . . . \left(i\right)$

Given $y\left(1\right) = 0 ⇒ \left(0\right)\left(1\right) = {e}^{1}\left[1 - 2\left(1\right) + 2\right] + C$

$⇒ C = - e$

$∴ y = \dfrac{{e}^{x}}{{x}^{2}}\left[{x}^{2} - 2x + 2\right] - e$ (from eq \left(i\right) $)Hence,$ z ( x ) = {x}^{2}\left(\dfrac{{e}^{x}}{{x}^{2}}\right)\left[{x}^{2} - 2x + 2\right] - e - {e}^{x} $$ = {e}^{x}\left[{x}^{2} - 2x + 2\right] - e - {e}^{x} $$ z\left(x\right) = {e}^{x}\left[{x}^{2} - 2x + 1\right] - e $$ {z}^{'}\left(x\right) = {e}^{x}\left(2x - 2\right) + \left({x}^{2} - 2x + 1\right){e}^{x} $$ {z}^{'}\left(x\right) = {e}^{x}\left[{x}^{2} - 1\right] $ To find local maxima, put$ {z}^{'}\left(x\right) = 0 $$ ⇒ {e}^{x}\left({x}^{2} - 1\right) = 0 ⇒ \left(x - 1\right)\left(x + 1\right) = 0 \left(∵ {e}^{x} > 0, \forall x \in R\right) $$ ⇒ x = 1, - 1 $<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/fac5d3d8-0ca7-4c9f-afbe-28e56422d7b3-image.png" style="width: 331px; height: 50px;" />It is clear from the sign scheme method,$ z\left(x\right) has local maximum value at $x = - 1$ and has local minimum value at $x = 1$ .

$∴$ The local maximum value is

$z\left(- 1\right)$ = {e}^{- 1} $\left[{\left(- 1\right)}^{2} - 2\left(- 1\right) + 1\right] - e$

$= \dfrac{1}{e}\left[1 + 2 + 1\right]$ - e = $\dfrac{4}{e} -$ e $$

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