Jee Main
Differential Equations
JEE Main 2021 (Online) 27th August Morning Shift
MCQ+4 / -12021
Let y = y(x) be the solution of the differential equation
\({{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x\) such that y(0) = 7. Then y(\(\pi\)) is equal to :
\({{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x\) such that y(0) = 7. Then y(\(\pi\)) is equal to :