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JEE Main 202227 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2022 · 27 June · Evening Shift · Mathematics · Differential Equations

Let y = y(x) be the solution of the differential equation (1 - x ^...

Let y=y(x)y = y\left(x\right) be the solution of the differential equation \left(1 - {x}^{2}\right)dy = \left(xy + \left({x}^{3} + 2\right)\sqrt{1 - {x}^{2}}\right)dx, - 1 < x < 1<br/>and<br />andy\left(0\right) = 0.If. If{∫}_{- \dfrac{1}{2}}^{\dfrac{1}{2}}\sqrt{1 - {x}^{2}}y\left(x\right)dx = kthenthen{k}^{- 1} is equal to

Correct answer

0

Step-by-step solution

Given,

(1−x2)dydx=xy+(x3+2)1−x2\left(1 - {x}^{2}\right)\dfrac{dy}{dx} = xy + \left({x}^{3} + 2\right)\sqrt{1 - {x}^{2}}

⇒dydx+(−x1−x2)y=x3+21−x2⇒ \dfrac{dy}{dx} + \left(\dfrac{- x}{1 - {x}^{2}}\right)y = \dfrac{{x}^{3} + 2}{\sqrt{1 - {x}^{2}}}

IF=e∫−x1−x2dx=1−x2IF = {e}^{∫ \dfrac{- x}{1 - {x}^{2}}dx} = \sqrt{1 - {x}^{2}}

y(x)⋅1−x2=x44+2x+cy\left(x\right) \cdot \sqrt{1 - {x}^{2}} = \dfrac{{x}^{4}}{4} + 2x + c

y(0)=0⇒c=0y\left(0\right) = 0 ⇒ c = 0

1−x2y(x)=x44+2x\sqrt{1 - {x}^{2}}y\left(x\right) = \dfrac{{x}^{4}}{4} + 2x

So,required value =∫−1212(x44+2x)dx−14⋅2∫012x4dx= {∫}_{- \dfrac{1}{2}}^{\dfrac{1}{2}}\left(\dfrac{{x}^{4}}{4} + 2x\right)dx - \dfrac{1}{4} \cdot 2{∫}_{0}^{\dfrac{1}{2}}{x}^{4}dx

=110(x5)012=1320= \dfrac{1}{10}{\left({x}^{5}\right)}_{0}^{\dfrac{1}{2}} = \dfrac{1}{320}

k−1=320{k}^{- 1} = 320

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