If (dy)/(dx) + e ^ x ( x ^ 2 - 2)y = ( x ^ 2 - 2x)( x ^ 2 - 2) e ^ 2x and y(0) = 0 , then the value of y(2) is

Correct answer: C. 0 Given (dy)/(dx) + e ^ x ( x ^ 2 - 2)y = ( x ^ 2 - 2x)( x ^ 2 - 2) e ^ 2x It is linear differential equation so, IF = e ^ ∫ ( x ^ 2 - 2) e ^ x dx = e ^ x ^ 2 e ^ x - 2 ∫ x e ^ x dx - 2 e ^ x = e ^ x ^ 2 e ^ x - 2[x e ^ x - e ^ x ] - 2 e ^ x IF = e ^ ( x ^ 2 - 2x) e ^ x Now solution is given by y × e ^ ( x ^ 2 - 2x) e ^ x = ∫ e ^ ( x ^ 2 - 2x) e ^ x × ( x ^ 2 - 2x)( x ^ 2 - 2) e ^ 2x = ∫ e ^ ( x ^ 2 - 2x) e ^ x × ( x ^ 2 - 2x) e ^ x ( x ^ 2 - 2) e ^ x Now let ( x ^ 2 - 2x) e ^ x = t We get e ^ x ( x ^ 2 - 2x) + e ^ x (2x - 2)dx = dt e ^ x ( x ^ 2 - 2x + 2x - 2)dx = dt e ^ x ( x ^ 2 - 2)dx = dt y × e ^ ( x ^ 2 - 2x) e ^ x = ∫ e ^ t × t × dt y × e ^ ( x ^ 2 - 2x) e ^ x = e ^ t (t - 1) + c y × e ^ ( x ^ 2 - 2x) e ^ x = e ^ ( x ^ 2 - 2x) e ^ x (( x ^ 2 - 2x) e ^ x - 1) + c Now at x = 0y = 0 0 × e ^ 0 = e ^ 0 (0x e ^ x - 1) + c 0 = - 1 + c c = 1 Putting the value of c = 1 we get the equation as ⇒ y × e ^ ( x ^ 2 - 2x) e ^ x = e ^ ( x ^ 2 - 2x) e ^ x (( x ^ 2 - 2x) e ^ x - 1) + 1 Now putting x = 2 in equation we get y × e ^ (4 - 4) e ^ 2 = e ^ (4 - 4) e ^ 2 ((4 - 4) e ^ 2 - 1) + 1 ⇒ y × e ^ 0 = e ^ 0 ((0) e ^ 2 - 1) + 1 ⇒ y = - 1 + 1 = 0

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 (dy)/(dx) + e ^ x ( x ^ 2 - 2)y = ( x ^ 2 - 2x)( x ^ 2 - 2) e ^...

\dfrac{dy}{dx} + {e}^{x}\left({x}^{2} - 2\right)y = \left({x}^{2} - 2x\right)\left({x}^{2} - 2\right){e}^{2x} and

y\left(0\right) = 0

, then the value of

y\left(2\right)$ is

Options

A $- 1$
B $1$
C $0$
D $e$

Correct answer

C. $0$

Step-by-step solution

Given

$\dfrac{dy}{dx} + {e}^{x}\left({x}^{2} - 2\right)y = \left({x}^{2} - 2x\right)\left({x}^{2} - 2\right){e}^{2x}$

It is linear differential equation so,

$IF = {e}^{∫ \left({x}^{2} - 2\right){e}^{x}dx}$

$= {e}^{{x}^{2}{e}^{x} - 2 ∫ x{e}^{x}dx - 2{e}^{x}} = {e}^{{x}^{2}{e}^{x} - 2\left[x{e}^{x} - {e}^{x}\right] - 2{e}^{x}}$

$IF = {e}^{\left({x}^{2} - 2x\right){e}^{x}}$

Now solution is given by

$y \times {e}^{\left({x}^{2} - 2x\right){e}^{x}} = ∫ {e}^{\left({x}^{2} - 2x\right){e}^{x}} \times \left({x}^{2} - 2x\right)\left({x}^{2} - 2\right){e}^{2x}$

$= ∫ {e}^{\left({x}^{2} - 2x\right){e}^{x}} \times \left({x}^{2} - 2x\right){e}^{x}\left({x}^{2} - 2\right){e}^{x}$

Now let \left({x}^{2} - 2x\right){e}^{x} = t $We get$ {e}^{x}\left({x}^{2} - 2x\right) + {e}^{x}\left(2x - 2\right)dx = dt $$ {e}^{x}\left({x}^{2} - 2x + 2x - 2\right)dx = dt $$ {e}^{x}\left({x}^{2} - 2\right)dx = dt $$ y \times {e}^{\left({x}^{2} - 2x\right){e}^{x}} = ∫ {e}^{t} \times t \times dt $$ y \times {e}^{\left({x}^{2} - 2x\right){e}^{x}} = {e}^{t}\left(t - 1\right) + c $$ y \times {e}^{\left({x}^{2} - 2x\right){e}^{x}} = {e}^{\left({x}^{2} - 2x\right){e}^{x}}\left(\left({x}^{2} - 2x\right){e}^{x} - 1\right) + c $Now at$ x = 0y = 0 $$ 0 \times {e}^{0} = {e}^{0}\left(0x{e}^{x} - 1\right) + c $$ 0 = - 1 + c c = 1 $Putting the value of$ c = 1 $we get the equation as$ ⇒ y \times {e}^{\left({x}^{2} - 2x\right){e}^{x}} = {e}^{\left({x}^{2} - 2x\right){e}^{x}}\left(\left({x}^{2} - 2x\right){e}^{x} - 1\right) + 1 $Now putting$ x = 2 $in equation we get$ y \times {e}^{\left(4 - 4\right){e}^{2}} = {e}^{\left(4 - 4\right){e}^{2}}\left(\left(4 - 4\right){e}^{2} - 1\right) + 1 $$ ⇒ y \times {e}^{0} = {e}^{0}\left(\left(0\right){e}^{2} - 1\right) + 1 $$ ⇒ y = - 1 + 1 = 0$

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