+ysec2x=sec2×tanx Given equation is linear differential equation. IF=e∫sec2xdx=etanx ⇒y.etanx=∫etanxsec2×tanxdx Put tanx=u=sec2xdx=du yetanx=∫euudu⇒yetanx=ueu−eu+c ⇒yetanx=(tanx−1)etanx+c ⇒y=(tanx−1)+c.e−tanx ∴y(0)=0 (given )⇒0=−1+c⇒c=1 Hence, solution of differential equation, y(−