Question

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

Answer 1

$(ax+b)e^{\frac{y}{x}}=x$

$e^{\frac{y}{x}}=\frac{x}{ax+b}$ ...$\left(1\right)$

Now differentiate on both sides

$\Rightarrow \frac{d}{dx}\left(e^{\frac{y}{x}}\right)=\frac{d}{dx}\left(\frac{x}{ax+b}\right)$

We use differentiation of $\frac{u}{v}$ formula

$d\frac{u}{v}=\frac{vdu-udv}{v^2}$

$e^{\frac{y}{x}}$ $\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{(ax+b)-x\left(a\right)}{ax+b^2}$

$e^{\frac{y}{x}}\frac{x\frac{dy}{dx}-y}{x^2}=\frac{ax+b-ax}{(ax+b{)}^2}$

This can be simplified as follows

$e^{\frac{y}{x}}\frac{x\frac{dy}{dx}-y}{x^2}=\frac{b}{(ax+b{)}^2}$ ...$\left(2\right)$

Now we replace the value of $e^{\frac{y}{x}}$ from $\left(1\right)$ in $\left(2\right)$

$\frac{x}{ax+b}$ $\frac{x\frac{dy}{dx}-y}{x^2}=\frac{b}{(ax+b{)}^2}$

Simplify by cancelling out the common terms

We'll be left with

$x\frac{dy}{dx}-y=\frac{bx}{ax+b}$ ....$\left(3\right)$

Now, differentiate again on both sides,

Use the formulae of $d\left(uv\right)$ and $d(\frac{u}{v}$

$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-\frac{dy}{dx}$ =$\frac{(ax+b)b-bx\left(a\right)}{(ax+b{)}^2}$

On simplifying this, we get

$x\frac{d^{2}y}{d{x}^2}=\frac{b^2}{(ax+b{)}^2}$

Multiply with $x^2$ on both sides

$x^3\frac{d^{2}y}{d{x}^2}=\frac{b^2{x}^2}{(ax+b{)}^2}$

$x^3\frac{d^{2}y}{d{x}^2}=(\frac{bx}{(ax+b)}{)}^2$

The value of $\frac{bx}{ax+b}$ from equation $\left(3\right)$ can be replaced in the above equation,

We get

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

Hence proved