dxdy+ycotx=sinx4x,x∈(0,π)
I.F. =e∫cotx=elnsinx=sinx
∴ysinx=∫sinx4x.sinxdx
ysinx=2x2+c
∵y(2π)=0
∴0×1=2×(2π)2+c
⇒c=−2π2
y(6π)=212×(6π)2−(2π2)=[(18π2)−(2π2)]×2
=9π2−π2
=−98π2
JEE Main 2018 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation sinxdxdy+ycosx=4x,x∈(0,π). If y(2π)=0, then y(6π) is equal to
Held on 8 Apr 2018 · Verified 6 Jul 2026.
−94π2
934π2
93−8π2
−98π2
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