We know that,
dxdy+y(2tanx)=sinx is a linear differential equation so,
I.F=e∫2tanxdx=e2ln∣secx∣=sec2x
The general solution will be
ysec2x=∫sinxsec2xdx+C
⇒ysec2x=secx+C
∵y(3π)=0⇒C=−2
Hence the particular solution is
ysec2x=secx−2
⇒y=cosx−2cos2x
⇒y=81−2(cosx−41)2
So, the maximum value of y(x) is 81