Q31: If $(a + bx)e^{\frac{y}{x}} = x$⁠, prove $x\frac{d^2y}{dx^2} = \left(\frac{a}{a + bx}\right)^2$

Divide both sides of $(a + bx)e^{\frac{y}{x}} = x$ by $a + bx$⁠. The aim is to free $y$ from the exponent, and a log does that next.

Take log on both sides and use two log rules:

Now there is no exponential left to differentiate.

Left side, by the quotient rule with $y$ on top and $x$ below:

Right side: $\frac{d}{dx}\log x = \frac1x$⁠, and by the chain rule $\frac{d}{dx}\log(a + bx) = \frac{b}{a + bx}$⁠.

For the chain rule, the inside is $a + bx$⁠, whose derivative is $b$⁠:

Multiply both sides by $x^2$⁠. Then $x^2 \cdot \frac1x = x$⁠, and $x^2\cdot\frac{b}{a + bx} = \frac{bx^2}{a + bx}$⁠.

Over the common denominator $a + bx$⁠:

Expand $x(a + bx) = ax + bx^2$⁠, then the $bx^2$ terms cancel:

Clearing the $x^2$ first keeps the second differentiation short.

Left side: $x\frac{dy}{dx}$ is a product, so its derivative is $x\frac{d^2y}{dx^2} + \frac{dy}{dx}\cdot 1$⁠. The derivative of $y$ is $\frac{dy}{dx}$⁠.

Right side, by the quotient rule:

with $u = ax$⁠, $u' = a$⁠, $w = a + bx$ and $w' = b$⁠.

The two $\frac{dy}{dx}$ terms cancel. On the right, $a^2 + abx - abx = a^2$⁠.

Finally $\frac{a^2}{(a + bx)^2} = \left(\frac{a}{a + bx}\right)^2$⁠, which is the required form.