Let $${\cos ^{ - 1}}\left( {{y \over b}} \right) = {\log _e}{\left( {{x \over n}} \right)^n}$$, then $$A{y_2} + B{y_1} + Cy = 0$$ is possible for, where $${y_2} = {{{d^2}y} \over {d{x^2}}},{y_1} = {{dy} \over {dx}}$$

The function $$y = {e^{kx}}$$ satisfies $$\left( {{{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}}} \right)\left( {{{dy} \over {dx}} - y} \right) = y{{dy} \over {dx}}$$. It is valid for

If $$y = {\log ^n}x$$, where $${\log ^n}$$ means $${\log _e}{\log _e}{\log _e}\,...$$ (repeated n times), then $$x\log x{\log ^2}x{\log ^3}x\,.....\,{\log ^{n - 1}}x{\log ^n}x{{dy} \over {dx}}$$ is equal to

If $$x = \sin \theta $$ and $$y = \sin k\theta $$, then $$(1 - {x^2}){y_2} - x{y_1} - \alpha y = 0$$, for $$\alpha=$$