Given,
(1−x2)dxdy=xy+(x3+2)1−x2
⇒dxdy+(1−x2−x)y=1−x2x3+2
IF=e∫1−x2−xdx=1−x2
y(x)⋅1−x2=4x4+2x+c
y(0)=0⇒c=0
1−x2y(x)=4x4+2x
So,required value =∫−2121(4x4+2x)dx−41⋅2∫021x4dx
=101(x5)021=3201
k−1=320
Let y=y(x) be the solution of the differential equation (1−x2)dy=(xy+(x3+2)1−x2)dx,−1<x<1
and y(0)=0. If ∫−21211−x2y(x)dx=k then k−1 is equal to
Held on 27 Jun 2022 · Verified 6 Jul 2026.
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