Question

The differential equation formed by eliminating a and b from the equation $y =e^x (a \cos x + b \sin x) $ is

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

Answer 1

Answer: (D)

Given,
$y=e^{x}(a \cos x+b \cos x) \dots$(i)
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=e^{x}(a \cos x+b \sin x)+e^{x}(-a \sin x+b \cos x)$
$\Rightarrow \frac{d y}{d x}=y+e^{x}(-a \sin x+b \cos x) \ldots$ (ii) [by Eq. (i)]
Again differentiating w.r.t. $x$, we get
$\frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}+e^{x}(-a \cos x-b \sin x)$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d y}{d x}-e^{x}(a \cos x+b \sin x)+\frac{d y}{d x}-y$
[By Eq. (ii) ]
$\Rightarrow \frac{d^{2} y}{d x^{2}}=2 \frac{d y}{d x}-y-e^{x}(a \cos x+b \sin x)$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=2 \frac{d y}{d x}-y-y [$ by Eq.(i)]
$\Rightarrow \frac{d^{2} y}{d x^{2}}=2 \frac{d y}{d x}-2 y$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+2 y-2 \frac{d y}{d x}=0$