Question

If $(ax+b)e^{y/x}=x$ or, $y=x\log \left(\frac{x}{a+bx}\right)$, prove that $x^3\frac{d^{2}y}{d{x}^2}={(x\frac{dy}{dx}-y)}^2$.

Answer 1

Given that,

$y=x\log \left(\frac{x}{a+bx}\right)$

Differentiation to x on both sides we have,

$\frac{dy}{dx}=x\times \frac{1}{\frac{x}{a+bx}}\times \frac{a+bx-xb}{{\left(a+bx\right)}^2}+\log \left(\frac{x}{a+bx}\right)$

$\frac{dy}{dx}=\left(a+bx\right)\times \frac{a}{{\left(a+bx\right)}^2}+\log \left(\frac{x}{a+bx}\right)$

$\frac{dy}{dx}=\frac{a}{\left(a+bx\right)}+\log \left(\frac{x}{a+bx}\right)$

Again differentiation to x on both sides we have

$\frac{dy}{dx}=x\times \frac{1}{\frac{x}{a+bx}}\times \frac{a+bx-xb}{{\left(a+bx\right)}^2}+\log \left(\frac{x}{a+bx}\right)$

$\frac{dy}{dx}=\left(a+bx\right)\times \frac{a}{{\left(a+bx\right)}^2}+\log \left(\frac{x}{a+bx}\right)$

$\frac{d^{2}y}{d{x}^2}=\frac{ad{\left(a+bx\right)}^{-1}}{d\left(a+bx\right)}\times \frac{d\left(a+bx\right)}{dx}+d\log \frac{\left(\frac{x}{a+bx}\right)}{d\left(\frac{x}{a+bx}\right)}\times \frac{d\left(\frac{x}{a+bx}\right)}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=a\left(-1\right){\left(a+bx\right)}^{-2}b+\frac{\frac{1}{x}}{a+bx}\times \frac{\left(a+bx-xb\right)}{{\left(a+bx\right)}^2}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-ab}{{\left(a+bx\right)}^{-2}}+\frac{\left(a+bx\right)}{x}\times \frac{a}{{\left(a+bx\right)}^2}$

$=\frac{-ab}{{\left(a+bx\right)}^{-2}}+\frac{a}{x\left(a+bx\right)}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-abx+a\left(a+bx\right)}{x+{\left(a+bx\right)}^2}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-abx+a^2+abx}{x+{\left(a+bx\right)}^2}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{a^2}{x+{\left(a+bx\right)}^2}$

$\therefore x^3\frac{d^{2}y}{d{x}^2}={\left(x\frac{dy}{dx}-y\right)}^2$