If dydx + ytanx = sin2x and y(0) = 1, then y(π) is equal to:

dydx+ytanx=sin2x
It is a linear equation
first we find integrating factor
I.F.=e∫tanxdx=secx (since integration of tanx=ln(secx))
Multiplying both sides by I.F.
dydxsecx+ytanxsecx=sin2xsecx
dydx(ysecx)=sin2xsecx
Integrating both sides∫d(ysecx)=∫sin2xsecxdx
ysecx=∫2sinxcosx1cosxdx
ysecx=∫2sinxdx
ysecx=-2cosx+c
y=-2cosxcos x+ccosx
Given y(0)=1
1=-2cos0cos0+ccos0
1=-2+c
c=3
Therefore, y=-2cosxcos x+3cosx
y(π)=-2cosπcosπ+3cosπ
=-2(-1)(-1)+3(-1)
=-2-3=-5