Question

If $y=\frac{u}{v}$, where u and v are functions of x,
then show
$v^3\frac{d^{2}y}{d{x}^2}=\left|\begin{array}{ccc}u& v& 0\\ u^{\prime }& v^{\prime }& v\\ u^{\prime \prime }& v^{\prime \prime }& 2v^{\prime }\end{array}\right|$

Answer 1

Since $y=\frac{u}{v}$

$\therefore \frac{dy}{dx}=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$

$\Rightarrow v^2\frac{dy}{dx}=v{u}^{\prime }-u{v}^{\prime }$

$=-\left|\begin{array}{cc}u& v\\ u^{\prime }& v^{\prime }\end{array}\right|$ ....... (1)

Differentiating both sides w.r.t. to x, we have

$v^2\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\cdot \left(2v{v}^{\prime }\right)=-\left|\begin{array}{cc}{u}^{\prime }& v^{\prime }\\ u^{\prime }& v^{\prime }\end{array}\right|-\left|\begin{array}{cc}u& v\\ u^{\prime \prime }& v^{\prime \prime }\end{array}\right|$

$=0-\left|\begin{array}{cc}u& v\\ u^{\prime \prime }& v^{\prime \prime }\end{array}\right|$

$\Rightarrow v^2\frac{d^{2}y}{d{x}^2}=-2v{v}^{\prime }\frac{dy}{dx}-\left|\begin{array}{cc}u& v\\ u^{\prime \prime }& v^{\prime \prime }\end{array}\right|$

or $v^3\frac{d^{2}y}{d{x}^2}=-2v^{\prime }{v}^2\frac{dy}{dx}-v\left|\begin{array}{cc}u& v\\ u^{\prime \prime }& v^{\prime \prime }\end{array}\right|$

$=2v^{\prime }\left|\begin{array}{cc}u& v\\ u^{\prime }& v^{\prime }\end{array}\right|-v\left|\begin{array}{cc}u& v\\ u^{\prime \prime }& v^{\prime \prime }\end{array}\right|$ {using (1)}

$=\left|\begin{array}{ccc}u& v& 0\\ u^{\prime }& v^{\prime }& v\\ u^{\prime \prime }& v^{\prime \prime }& 2v^{\prime }\end{array}\right|$

Hence
$v^3\frac{d^{2}y}{d{x}^2}=\left|\begin{array}{ccc}u& v& 0\\ u^{\prime }& v^{\prime }& v\\ u^{\prime \prime }& v^{\prime \prime }& 2v^{\prime }\end{array}\right|$