Let y=f(x) be the solution of the differential equation d y ~d x +(x y)/(x^2-1)= x^6+4 x 1-x^2 ,-1 x 1 such that f(0)=0 . If 6 ∫_ -1 / 2 ^ 1 / 2 f(x) d x=2 π-α then α^2 is equal to _______ .

Correct answer: 0 I.F. e^ -(1)/(2) ∫ 2 x 1-x^2 dx =e^ -(1)/(2) (1-x^2) = 1-x^2 y × 1-x^2 =∫(x^6+4 x) d x=(x^7)/(7)+2 x^2+c Given y(0)=0 c=0 y= (x^7)/(7)+2 x^2 1-x^2 Now, 6 ∫_ -(1)/(2) ^ (1)/(2) (x^7)/(7)+2 x^2 1-x^2 dx=6 ∫_ -(1)/(2) ^ (1)/(2) 2 x^2 1-x^2 dx aligned & =24 ∫_0^ (1)/(2) x^2 1-x^2 d x & Put x= θ & d x= θ d θ & =24 ∫_0^ (π)/(6) ( ^2 θ)/( θ) θ d θ & =24 ∫_0^ (π)/(6) ((1- 2 θ)/(2)) d θ=12[θ-( 2 θ)/(2)]_0^ (π)/(6) & =12((π)/(6)- 3 4 ) & =2 π-3 3 & α^2=(3 3 )^2=27 aligned

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Differential Equations

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

Let

y=f(x)

be the solution of the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}},-1 \lt x \lt 1

such that

f(0)=0

. If

6 \int_{-1 / 2}^{1 / 2} f(x) \mathrm{d} x=2 \pi-\alpha

then

\alpha^2

is equal to _______ .

Correct answer

Step-by-step solution

I.F.

\mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2} y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c

Given

y(0)=0 \Rightarrow c=0 y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}}

Now,

6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7}+2 \mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx}=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 \mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx}

\begin{aligned} & =24 \int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} d x \\ & \text{ Put } x=\sin \theta \\ & d x=\cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\ & =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\ & =2 \pi-3 \sqrt{3} \\ & \alpha^2=(3 \sqrt{3})^2=27\end{aligned}

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