Question

If $y = e^{-x}\, cos2x$ then which of the following differential equations is satisfied ?

• A. $\frac{d^{2}y}{dx^{2}}+2 \frac{dy}{dx}+5y=0$
• B. $\frac{d^{2}y}{dx^{2}}+5 \frac{dy}{dx}+2y=0$
• C. $\frac{d^{2}y}{dx^{2}}-5 \frac{dy}{dx}-2y=0$
• D. $\frac{d^{2}y}{dx^{2}}+2 \frac{dy}{dx}-5y=0$

Answer 1

Answer: (A)

Given, $y=e^{-x} \cos 2 x$ ...(i)
$\therefore \frac{d y}{d x}=e^{-x}(-\sin 2 x) 2+\cos 2 x \cdot e^{-x}(-1)$
$\Rightarrow \frac{d y}{d x}=-2 \sin 2 x \cdot e^{-x}-y$
$\Rightarrow \frac{d y}{d x}+y=-2 \sin 2 x \cdot e^{-x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}$
$=-2\left[\sin 2 x \cdot e^{-x}(-1)\right.$
$\left.+ e^{-x} 2 \cos 2 x\right]$
$=2 \sin 2 x \cdot e^{-x}-4 y$ [from Eq. (i)]
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\frac{d y}{d x}-y-4 y$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$