Q.

Let y = f(x) be the solution of the differential equation dydx+xyx2–1=x6+4x1–x2, – 1 < x < 1 such that f(0) = 0. If 6∫–1/21/2f(x)dx=2π–α, then α2 is equal to __________.          [2025]


Ans.

(27)

From given differential equation, we have

I.F.=e–12∫2x1–x2dx=e12ln(1–x2)=1–x2

Hence, solution of given D.E. is

y×1–x2=∫(x6+4x)dx=x77+2x2+c

Given, y(0) = 0, then c = 0

∴  y=x77+2x21–x2

Let  I=6∫–1212x77+2x21–x2dx          ... (i)

⇒ I=6∫–1212–x77+2x21–x2dx          ... (ii)

Adding (i) & (ii), we get

2I=24∫–1212x21–x2dx=2×24∫012x21–x2dx

Put x=sinθ ⇒ dx=cosθdθ

∴  I=24∫0π6sin2θcosθcosθdθ

=24∫0π6(1–cos2θ2)dθ=12[θ–sin2θ2]0π6

=12(π6–34)=2π–33

On comparing, we get α2=(33)2=27.