Question

If $y=e^{ax}\sin bx$, then$\frac{d^{2}y}{d{x}^2}-2a\frac{dy}{dx}+a^{2}y$ is equal to?

• A. $0$
• B. $1$
• C. $-b^{2}y$
• D. $-by$

Answer 1

The correct option is C

$-b^{2}y$

Explanation for the correct answer:

Step 1: Find the value of $\frac{dy}{dx}$

Given, $y=e^{ax}\sin bx$

Now, differentiate the given function with respect to $x$, we get;

$$\begin{gathered} \frac{dy}{dx}=a{e}^{ax}\sin bx+b{e}^{ax}\cos bx\left[\because \frac{d\left(u·v\right)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx},\frac{d{e}^x}{dx}=e^x,\frac{d\left(\sin x\right)}{dx}=\cos x\right] \\ \Rightarrow \frac{dy}{dx}=ay+b{e}^{ax}\cos bx \\ \Rightarrow \frac{dy}{dx}–ay=b{e}^{ax}\cos bx \end{gathered}$$

Step 2: Find the value of $\frac{d^{2}y}{d{x}^2}-2a\frac{dy}{dx}+a^{2}y$

Again, differentiate obtained function with respect to $x$

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}–a\frac{dy}{dx}=ab{e}^{ax}\cos bx–b^2{e}^{ax}\sin bx\left[\because \frac{d\left(u·v\right)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx},\frac{d{e}^x}{dx}=e^x,\frac{d\left(\cos x\right)}{dx}=-\sin x\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}–a\frac{dy}{dx}=a\left(\frac{dy}{dx}–ay\right)–b^{2}y \\ \Rightarrow \frac{d^{2}y}{d{x}^2}–2a\frac{dy}{dx}+a^{2}y=–b^{2}y \end{gathered}$$

Hence, the correct option is C.