We have $r$ and $\theta$ are polar coordinates in 2D and $h = r^2 \dot \theta$. Verify that $$ u(\theta) = \frac{\sqrt{\theta^2+c^4}}{ac} $$ where $c$ is any constant solves $$ \frac{d^2u}{d\theta^2} = \frac{1}{a^4u^3}. $$
I initially tried differentiating $u(\theta)$ twice and seeing if it was equal to $\dfrac{1}{a^4u^3}$, but this didn't work.
My working: I have worked out by differentiating $u$ twice that $$ \frac{d^2u}{d\theta^2} = \frac{c^3}{a(\theta^2+c^4)^\frac{3}{2}}. $$
Subbing in for $u$, I get that $$ \frac{1}{a^4u^3}=\frac{c^3}{a(\theta^2+c^4)^\frac{3}{2}}. $$
Since $$ \frac{d^2u}{d\theta^2} = \frac{c^3}{a(\theta^2+c^4)^\frac{3}{2}} = \frac{c^3}{a(\theta^2+c^4)^\frac{3}{2}}, $$ $u(\theta)$ is a solution of the second order differential equation stated.
Is this correct ?