Question

If $y=A\cos nx+B\sin nx$, then $\frac{d^{2}y}{d{x}^2}$ equals to?

• A. $n^{2}y$
• B. $-y$
• C. None of these
• D. $-n^{2}y$

Answer 1

The correct option is D

$-n^{2}y$

Explanation for the correct option:

Finding the double deritvative:

Given that,

$y=A\cos nx+B\sin nx...\left(1\right)$

Differentiate the above equation with respect to $x$,

$\frac{dy}{dx}=-An\sin nx+Bn\cos nx\left[\because \frac{d\left(cosax\right)}{dx}=-asinx,\frac{d\left(sinax\right)}{dx}=acosx\right]$

Differentiate again with respect to $x$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& -A{n}^2\cos nx–B{n}^2\sin nx\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{cosax}\mathbf{\right)}}{\mathbf{dx}}\mathbf{=}\mathbf{-}\mathbf{asinx}\mathbf{,}\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{sinax}\mathbf{\right)}}{\mathbf{dx}}\mathbf{=}\mathbf{acosx}\mathbf{]}\\ & =& -n^2(A\cos nx+B\sin nx)\\ & =& -n^{2}y\left[\mathrm{from}\left(1\right)\right]\end{array}$

Hence, the correct option is D.