Question

If $y=A\sin x+B\cos x$, then prove that $\frac{d^{2}y}{d{x}^2}+y=0$.

Answer 1

Differentiate $y=A\sin x+B\cos x$ with respect to $x$.

$\frac{dy}{dx}=A\cos x-B\sin x$

Again differentiate with respect to $x$.

$\frac{d^{2}y}{d{x}^2}=-A\sin x-B\cos x$

$\frac{d^{2}y}{d{x}^2}=-\left(A\sin x+B\cos x\right)$

$\frac{d^{2}y}{d{x}^2}=-y$

$\frac{d^{2}y}{d{x}^2}+y=0$