Question

If $xy=A\sin x+B\cos x$ is the solution of the differential equation $x\frac{d^{2}y}{d{x}^2}-5a\frac{dy}{dx}+xy=0$, then the value of $a$ is

• A. $\frac{2}{5}$
• B. $\frac{5}{2}$
• C. $\frac{-2}{5}$
• D. $\frac{-5}{2}$
• E. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{-2}{5}$

Given,
$xy=A\sin x+B\cos x$ .... (i)
On differentiating w.r.t. $x$, two times, we get
$x\frac{dy}{dx}+y=A\cos x-B\sin x$ .... (ii)
and $x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{dy}{dx}=-A\sin x-B\cos x$
$\Rightarrow x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}=-xy$ [from Eq. (i)]
$\Rightarrow x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+xy=0$
On Comparing with $x\frac{d^{2}y}{d{x}^2}-5a\frac{dy}{dx}+xy=0$, we get
$-5a=2\Rightarrow a=-\frac{2}{5}$