If x = tan(1/alog y) then show that (1 + x²)(d²y)/(dx²) + (2x - a)(dy)/(dx) = 0

If $x = \tan\left(\frac{1}{a}\log y\right)$ then show that $(1 + x^2)\frac{d^2y}{dx^2} + (2x - a)\frac{dy}{dx} = 0$

Given: $x = \tan\left(\frac{1}{a} \log y\right)$

To show: $(1 + x^2)\frac{d^2 y}{dx^2} + (2x - a)\frac{dy}{dx} = 0$

Answer

Proof of the differential equation

1. $x=\tan\left(\frac1a\log y\right)$ gives $\log y=a\tan^{-1}x$.
2. Differentiate with respect to $x$: $\frac1y\frac{dy}{dx}=\frac{a}{1+x^2}$.
3. Hence $(1+x^2)\frac{dy}{dx}=ay$.
4. Differentiate again (product rule): $(1+x^2)\frac{d^2y}{dx^2}+2x\frac{dy}{dx}=a\frac{dy}{dx}$.
5. Hence $(1+x^2)\frac{d^2y}{dx^2}+(2x-a)\frac{dy}{dx}=0$.