If x = tan(1/a log y), prove that (1 + x²)(d²y)/(dx²) + (2x - a)(dy)/(dx) = 0

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

Answer

Proof that $(1 + x^2)\frac{d^2y}{dx^2} + (2x - a)\frac{dy}{dx} = 0$.

Given $x = \tan\left(\frac{1}{a}\log y\right)$, so $\log y = a\tan^{-1}x$.
Differentiate with respect to $x$: $\frac{1}{y}\frac{dy}{dx} = \frac{a}{1 + x^2}$.
Hence $(1 + x^2)\frac{dy}{dx} = ay$.
Differentiate again (product rule): $2x\frac{dy}{dx} + (1 + x^2)\frac{d^2y}{dx^2} = a\frac{dy}{dx}$.
Rearranging: $(1 + x^2)\frac{d^2y}{dx^2} + (2x - a)\frac{dy}{dx} = 0$.
Hence proved.