Question

If $(ax+b)c\frac{y}{x}=x$, then show that $X^3\left(\frac{d^{2}y}{d{x}^2}\right)={(x\frac{dy}{dx}-y)}^2$

Answer 1

We have $x=(ax+b)e\frac{y}{x}$ $\Rightarrow logx=log\left[\right(ax+b\left)e\frac{y}{x}\right].$

$\Rightarrow logX=log(ax+b)+loge\frac{y}{x}$ $\Rightarrow logx-log(ax+b)=\frac{y}{x}loge$

or y = $xlog\left(\frac{x}{ax+b}\right)$ $\Rightarrow y=xlogx-log(ax+b)....\left(A\right)$

$\Rightarrow \frac{dx}{dy}=x[\frac{1}{x}-\frac{a}{ax+b}]+[logx-log(ax+b\left)\right].1$ [By (A), $\frac{y}{x}=logx-log(ax+b)]$

$\left[\begin{array}{ccc}\frac{dy}{dx}=\frac{b}{ax+b}+\frac{y}{x}\dots \left(i\right)& \Rightarrow xy=\frac{bx}{ax+b}+y& \\ \Rightarrow xy+y=\frac{(ax+b)b-bx\left(a\right)}{(ax-b{)}^2}& \Rightarrow xy=\frac{b^2}{(ax+b{)}^2}={(y-\frac{y}{x})}^2& \left[By\right(i)y-\frac{y}{x}=\frac{b}{ax+b}]\\ Thatis,x{y}^{\prime \prime }=\frac{(x{y}^{\prime }-y{)}^2}{x^2}& \Rightarrow x^{3}y=(xy-y{)}^2& \therefore x^3\frac{d^{2}y}{d{x}^2}={(x\frac{dy}{dx}-y)}^2\end{array}\right]$