Browse Questions

# The differential equation for which $y=a\cos x+b\sin x$ is a solution is $(A)\;\frac{d^2y}{dx^2}+y=0 \quad (B)\;\frac{d^2y}{dx^2}-y=0 \quad (C)\;\frac{d^2y}{dx^2}+(a+b)y=0 \quad (D)\;\frac{d^2y}{dx^2}+(a+b)y=0$

Toolbox:
• A differential equation will be of $n^{th}$ order if its equation has $'n'$ arbitrary constants
• $\large\frac{d}{dx}$$(\sin x)=\cos x and \large\frac{d}{dx}$$(\cos x)=-\sin x$
Given $y=a \cos x+b \sin x$
Differentiate w.r.t x on both sides,
$\large\frac{dy}{dx}$$=-a\sin x+b \cos x Again differentiate w.r.t x on both sides, \large\frac{d^2y}{dx^2}$$=-a\cos x-b \sin x$
$=-[a \cos x+b\sin x]$
But $a \cos x+b \sin x=y$
Hence $\large\frac{d^2y}{dx^2}=-y$
$\large\frac{d^2y}{dx^2}+y=0$
Hence the correct option is $A$