+esinycosx=cosx . . . (i) Let esiny=t, then esiny⋅cosy⋅
dy
dx
=
dt
dx′
Putting in Eq. (i), cosx ... (ii) (Linear form) Then, IF =e∫cosxdx=esinx Solution of differential Eq. (ii) is, t⋅IF=∫cosx⋅IFdx+C t⋅esinx=∫cosx⋅esinxdx+C =eu i.e. let sinx=u then cosxdx=du ⇒t⋅esinx=∫eudu+C=eu+C Put u=sinx and t=esiny ⇒esiny⋅esinx=esinx+C Given, y(0)=0, this gives C=0 ⇒esiny⋅esinx=esinx ⇒esiny+sinx=esinx ⇒siny+sinx=sinx ⇒siny=0 ⇒y=0 ∴y(π∕6)=y(π∕3)=y(π∕4)=0 Hence, 1+y(