Question

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

Answer 1

$\begin{array}{c}\left(a+bx\right)e^{y/x}=x\\ e^{y/x}=\frac{x}{a+bx}\\ taking\log \\ \log e^{y/x}=\log \left(\frac{x}{a+bx}\right)\\ \frac{y}{x}=\log x-\log \left(a+bx\right)\\ y=x\left[\log x-x\log \left(a+bx\right)\right]\\ y^{\prime }=x.\frac{1}{x}+\log x.1-x\left(\frac{1}{a+bx}\right)\times b-\log \left(a+bx\right)\to \left(i\right)\\ y^{\prime \prime }\frac{1}{x}-b\left\{\frac{-\left(a+bx\right)\times 1-x\left(b\right)}{{\left(a+bx\right)}^2}-\frac{1}{a+bx}\times b\right\}\\ x^3{y}^{\prime \prime }={\left(x\frac{dy}{dx}\right)}^2-2xy\frac{dy}{dx}+y^2\\ ={\left(x\frac{dy}{dx}-y\right)}^2\end{array}$

Proved.