dxdy+2tanx⋅y=2sinx
I.F. =e∫2tanxdx=sec2x
solution is
y⋅sec2x=∫2sinx⋅sec2xdx+C
ysec2x=2secx+C
⇒0=2.2+C⇒C=−4
ysec2x=2secx−4
y(4π)=2−2
JEE Main 2020 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation cosxdxdy+2ysinx=sin2x,x∈(0,2π)
If y(π/3)=0, then y(π/4) is equal to :
Held on 5 Sept 2020 · Verified 6 Jul 2026.
2−2
2+2
2−2
21−1
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