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If $y=e^{-x} \cos 2 x$, then which of the following differential equation is satisfied?
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Verified Answer
The correct answer is:
$\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$
Given, $y=e^{-x} \cos 2 x$
$\therefore \quad \frac{d y}{d x}=e^{-x}(-\sin 2 x) 2+\cos 2 x \cdot e^{-x}(-1)$
$\Rightarrow \quad \frac{d y}{d x}=-2 \sin 2 x \cdot e^{-x}-y$
$\Rightarrow \quad \frac{d y}{d x}+y=-2 \sin 2 x \cdot e^{-x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-2\left[\sin 2 x \cdot e^{-x}(-1)\right. \left.e^{-x} 2 \cos 2 x\right]$
$=2 \sin 2 x \cdot e^{-x}-4 y \quad$ |from Eq. $\left.(0)\right]$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\frac{d y}{d x}-y-4 y$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$
$\therefore \quad \frac{d y}{d x}=e^{-x}(-\sin 2 x) 2+\cos 2 x \cdot e^{-x}(-1)$
$\Rightarrow \quad \frac{d y}{d x}=-2 \sin 2 x \cdot e^{-x}-y$
$\Rightarrow \quad \frac{d y}{d x}+y=-2 \sin 2 x \cdot e^{-x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-2\left[\sin 2 x \cdot e^{-x}(-1)\right. \left.e^{-x} 2 \cos 2 x\right]$
$=2 \sin 2 x \cdot e^{-x}-4 y \quad$ |from Eq. $\left.(0)\right]$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\frac{d y}{d x}-y-4 y$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$
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