+ytanx=sin2x I.F=e∫tanxdx=e−logcosx=secx Required solution is y(secx)=∫sin2xsecxdx+c y(secx)=∫
2sinxcosx
cosx
dx+c y(secx)=2∫sinxdx+c y(secx)=−2cosx+c .........(1) Given y(0)=1 ∴ put x=0 and y=1, we get 1(sec0)=−2cos0+c ⇒c=1+2⇒c=3 ∴ from eqn (1), we have ysecx=−2cosx+3 To find y(π), put x=π in eqn (2), we get y(secπ)=−2cosπ+3 y=−2(−1)(−1)+3(−1)=−2−3=−5