The particular solution of the differential equation sin−1(d2ydx2−1)=x, wherey=dydx=0 whenx=0, is

[b] The differential equation is d2ydx2=1+sinx                          ?. (i) Integrating we get dydx=x−cosx+c     ? (ii) When x=0,dydx=0⇒c=1 ∴ Equation (ii) is dydx=x−cosx+1 Integrating again we get y=x22−sinx+x+D                         ? (iii) When x=0,y=0⇒D=0 ∴ The particular solution is y=x22+x−sinx