Q.

If y = y(x) is the solution of the differential equation 4–x2dydx=((sin–1(x2))2–y)sin–1(x2), –2≤x≤2, y(2)=π2–84, then y2(0) is equal to _________.         [2025]


Ans.

(4)

From the given D.E.

dydx+(sin–1x2)4–x2y=(sin–1x2)34–x2

I.F.=e∫sin–1x24–x2dx=e(sin–1x2)22

∴   ye(sin–1x2)22=∫(sin–1x2)34–x2e(sin–1x2)22dx

⇒ y=(sin–1x2)2–2+ce–(sin–1x2)22

Now, y(2)=π24–2+ce–π28

On comparing with y(2)=π2–84, we get c = 0

∴  y=(sin–1x2)2–2

Hence, y(0)=–2 ⇒y2(0)=4