Question

The differential equation whose solution represents the family $xy = Ae^{ax} + Be^{-ax}$ is

• A. $x\left(\frac{d^{2}\,y}{dx^{2}}\right)^{2}+2 \frac{dy}{dx}=xy$
• B. $x\left(\frac{d^{2}\,y}{dx^{2}}\right)^{2}+2 \frac{dy}{dx}=a^{2}xy$
• C. $x\frac{d^{2}\,y}{dx^{2}}+2 \frac{dy}{dx}=a^{2}xy$
• D. none of these

Answer 1

Answer: (C)

Given $xy = Ae^{ax} + Be^{-ax}\quad \ldots (i)$
Differentiating $(i)$ w.r.t. $x$, we get
$y+x \frac{dy}{dx}=a\left(Ae^{ax}-Be^{-ax}\right)$
Differentiating again w.r.t. $x$, we get
$\frac{dy}{dx}+x \frac{d^{2}\,y}{dx^{2}}+\frac{dy}{dx}=a^{2}\left(Ae^{ax}+Be^{-ax}\right)$
$\Rightarrow x\frac{d^{2}\,y}{dx^{2}}+2\frac{dy}{dx}=a^{2}\left(xy\right)\,\,\,$ (using $(i)$)