We have
dzdx=12sec2(x/2)tan(x/2)=1sinx=cosecx
dydx=dydz.dzdx=cosecxdydz
d2ydx2=ddx(dydx)=ddz(dydx)dzdx
=ddz(cosecxdydz)cosecx
=(cosecxd2ydx2−cosecxcotxdxdzdydz)cosecx
=cosec2xd2ydz2−cosecxcotxdydz
Putting these values in the given differential equation, we have
d2ydx2+cotxdydx+4ycosec2x=0
⇔ cosec2xd2ydx2−cosecxcotxdydz+cotxcosecxdydz+4ycosec2x=0
⇔ cosec2x(d2ydz2+4y)=0 ⇔ d2ydz2+4y=0
⇒A=4