Question

Let $y=a\cos \omega x+b\sin \omega x$ then prove that $\frac{d^{2}y}{d{x}^2}+{\omega }^{2}x=0$.

Answer 1

Given,

$y=a\cos \omega x+b\sin \omega x$

Now differentiating both sides w.r.to $x$ we get,

$\frac{dy}{dx}=\omega (-a\sin \omega x+b\cos \omega x)$

Now again differentiating both sides w.r.to $x$ we get,

$\frac{d^{2}y}{d{x}^2}={\omega }^2(-a\cos \omega x-b\sin \omega x)$

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

or, $\frac{d^{2}y}{d{x}^2}+{\omega }^{2}y=0$.