Question

If $y=e^{ax}\cos bx$, then find the value of $\frac{d^{2}y}{d{x}^2}-2a\frac{dy}{dx}+(a^2+b^2)y$.

• A. $0$
• B. $1$
• C. $e^{ax}$
  
• D. $3ab$

Answer 1

The correct option is A $0$

Let $y=e^{ax}\cos bx$ .....(1)
Differentiate both sides with respect to '$x$'
$\frac{dy}{dx}=e^{ax}(-b\sin bx)+a{e}^{ax}\cos bx$
$\frac{dy}{dx}=e^{ax}\left[a\cos bx-b\sin bx\right]$ ....(2)

Again differentiating, we get
$\frac{d^{2}y}{d{x}^2}=a{e}^{ax}\left[a\cos bx-b\sin bx\right]+e^{ax}[-ab\sin bx-b^2\cos bx]$
$\frac{d^{2}y}{d{x}^2}=a{e}^{ax}\left[a\cos bx-b\sin bx-ab\sin bx-b^2\cos bx\right]$
$\frac{d^{2}y}{d{x}^2}-2a\frac{dy}{dx}+(a^2+b^2)y$
$=e^{ax}\left[\cos bx(a^2-b^2)-\sin bx\left(2ab\right)\right]-2a{e}^{ax}[a\cos bx-b\sin bx]+\left(a^2{+b}^2\right)e^{ax}\cos bx$
$=e^{ax}[a^2\cos bx-b^2\cos bx-2ab\sin bx-2a^2\cos bx+2ab\sin bx+a^2\cos bx+b^2\cos bx]$
$=e^{ax}\left(0\right)$
$=0$