Jee Main
Differential Equations
JEE Main 2021 (Online) 25th July Evening Shift
MCQ+4 / -12021
Let y = y(x) be the solution of the differential
equation xdy = (y + x3 cosx)dx with y(\(\pi\)) = 0, then \(y\left( {{\pi \over 2}} \right)\) is equal to :
equation xdy = (y + x3 cosx)dx with y(\(\pi\)) = 0, then \(y\left( {{\pi \over 2}} \right)\) is equal to :