Question

If $y=5cosx-3sinx$,prove that $\frac{d^2y}{dx^2}+y=0$

Answer 1

It is given that,$y=5cosx-3sinx$
Then,
$\frac{dy}{dx}=\frac{d}{dx}(5cosx)-\frac{d}{dx}(3sinx)=5\frac{d}{dx}(cosx)-3\frac{d}{dx}(sinx)$
$=5(-sinx)-3cosx=-(5sinx+3cosx)$
$∴\frac{d^2y}{dx^2}=\frac{d}{dx}[-(5sinx+3cosx)]$
$=-[5.\frac{d}{dx}(sinx)+3.\frac{d}{dx}(cosx)]$
$=-[5cosx+3(-sinx)]$
$=-[5cosx-3sinx]$
$=-y$
$∴\frac{d^2y}{dx^2}+y=0$
Hence, proved.