Question

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

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

Answer 1

Answer: (D)

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