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If $\left ( a+bx \right )e^{\frac{y}{x}}= x,$ then prove that
$x^{3}\frac{d^{2y}}{dx^{2}}= \left ( x\frac{dy}{dx}-y \right )^{2}$

Answers (1)

we have, $\left ( a+bx \right )e^{\frac{y}{x}}= x$
$\Rightarrow e^{\frac{y}{x}}= \frac{x}{a+bx}$
$\Rightarrow \frac{y}{x}= \log \left ( \frac{x}{a+bx} \right )\Rightarrow y=x\log \left ( \frac{x}{a+bx} \right )$
$\Rightarrow y= x\left \{ \log x-\log \left ( a+bx \right ) \right \} \Rightarrow \frac{y}{x}= \log x-\log \left ( a+bx \right )$ On differentiating w.r.t x, we get
$\frac{x\frac{dy}{dx}-y}{x^{2}}= \frac{1}{x}-\frac{1}{a+bx}xb$ $\Rightarrow \frac{1}{x^{2}}\left ( \frac{xdy}{dx}-y \right )= \frac{1}{x}-\frac{b}{a+bx}$
$\Rightarrow x\frac{dy}{dx}-y= x^{2}\left ( \frac{1}{x}-\frac{b}{a+bx} \right )$
$\Rightarrow x\frac{dy}{dx}-y= \frac{ax}{a+bx}-(i)$
Again,differentiating both side w.r.t x
$\frac{xd^{2}y}{dx^{2}}+\frac{dy}{dx}-\frac{dy}{dx}= \frac{\left ( a+bx \right )a-ax\times b}{\left ( a+bx \right )^{2}}$
$\Rightarrow \frac{xd^{2}y}{dx^{2}}= \frac{a^{2}+abx-abx}{\left ( a+bx \right )^{2}}$
$\Rightarrow \frac{xd^{2}y}{dx^{2}}= \frac{a^{2}}{\left ( a+bx \right )^{2}}$
On multiplying both side by $x^{2}$ ,we get
$x^{3}\frac{d^{2}y}{dx^{2}}= \frac{a^{2}x^{2}}{\left ( a+bx \right )^{2}}$
$x^{3}\frac{d^{2}y}{dx^{2}}=\left ( \frac{ax}{a+bx} \right )^{2}-(ii)$
from (i) and(ii) we get
$x^{3}\frac{d^{2}y}{dx^{2}}=\left ( \frac{xdy}{dx} -y\right )^{2}$
hence proved.