Let y = y ( x ) be the solution of the differential equation (dy)/(dx) = 2(y + 2 x - 5)x - 2 x such that y ( 0 ) = 7 . Then y ( π ) is equal to

Correct answer: C. 2 e ^ π ^ 2 + 5 (dy)/(dx) - 2xy = 2(2 x - 5)x - 2 x IF = e ^ - x ^ 2 So, y e ^ - x ^ 2 = ∫ e ^ - x ^ 2 (2x(2 x - 5) - 2 x)dx ⇒ y · e ^ - x ^ 2 = e ^ - x ^ 2 (5 - 2 x) + C y = 5 - 2 x + C · e ^ x ^ 2 Given at x = 0, y = 7 ⇒ 7 = 5 + C ⇒ C = 2 So, y = 5 - 2 x + 2 e ^ x ^ 2 Now at x = π y = 5 + 2 e ^ π ^ 2

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Differential Equations

Let y = y ( x ) be the solution of the differential equation...

Let

y = y ( x )

be the solution of the differential equation

\dfrac{dy}{dx} = 2\left(y + 2\sin x - 5\right)x - 2\cos x

such that

y ( 0 ) = 7 .

Then

y ( \pi )

is equal to

Options

A $7{e}^{{\pi}^{2}} + 5$
B ${e}^{{\pi}^{2}} + 5$
C $2{e}^{{\pi}^{2}} + 5$
D $3{e}^{{\pi}^{2}} + 5$

Correct answer

C. $2{e}^{{\pi}^{2}} + 5$

Step-by-step solution

$\dfrac{dy}{dx} - 2xy = 2\left(2\sin x - 5\right)x - 2\cos x$

$IF = {e}^{- {x}^{2}}$

So, $y{e}^{- {x}^{2}} = ∫ {e}^{- {x}^{2}}\left(2x\left(2\sin x - 5\right) - 2\cos x\right)dx$

$⇒ y \cdot {e}^{- {x}^{2}} = {e}^{- {x}^{2}}\left(5 - 2\sin x\right) + C$

$y = 5 - 2\sin x + C \cdot {e}^{{x}^{2}}$

Given at $x = 0, y = 7$

$⇒ 7 = 5 + C$

$⇒ C = 2$

So, $y = 5 - 2\sin x + 2{e}^{{x}^{2}}$

Now at $x = \pi$

$y = 5 + 2{e}^{{\pi}^{2}}$

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