(1+y2)etanxdx+cos2x(1+e2tanx)dy=0∫1+e2tanxsec2xetanxdx+∫1+y2dy=C⇒tan−1(etanx)+tan−1y=C for x=0,y=1,tan−1(1)+tan−11=C C=2π tan−1(etanx)+tan−1y=2π
Put x=π,tan−1e+tan−1y=2π tan−1y=cot−1e y=e1
JEE Main 2024 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation (1+y2)etanxdx+cos2x(1+e2tanx)dy=0,y(0)=1. Then y(4π) is equal to
Held on 8 Apr 2024 · Verified 6 Jul 2026.
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