Question

The relation $y=A\sin x+B\cos x$ can be represented by the differential equation

• A. $\frac{dy}{dx}=y$
• B. $\frac{d^{2}y}{d{x}^2}=y$
• C. $\left(\frac{d^{2}y}{d{x}^2}\right)+y=0$
• D. None of these

Answer 1

The correct option is C

$\left(\frac{d^{2}y}{d{x}^2}\right)+y=0$

Explanation for the correct option:

Write the differential equation of the given equation.

Given, $y=A\sin x+B\cos x$

Differentiating w.r.t. $x$, we get

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

Again differentiating w.r.t. $x$, we get

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

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

We know that $y=A\sin x+B\cos x$

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

$\Rightarrow \left(\frac{d^{2}y}{d{x}^2}\right)+y=0$

Hence, option $C$ is the correct answer.