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

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

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