STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

Differentiate both sides with respect to $x$, we get

$ - y\sin x + \cos x.y' + x( - \sin y)y' + \cos y$

Again differentiate with respect to $x$

$ - y''\sin x - y\cos x + \cos x.y'' + \sin x.y' - \sin y.y'$

$ - x[\cos y.{(y')^2} + \sin y.y''] - \sin y.y'$

Putting $x = 0$, we get $ - y + y'' - 2\sin y\,y' = 0$

$y'' = y + 2y'\sin y$

Since at $x = 0$, $y = \pi $; ${(y'')_0} = \pi $.