Question

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

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

Answer 1

Answer: (C)

$\frac{-2}{5}$

Answer 2

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