Locus Equation $f(r) = \frac{-h^2}{r^3}$?

Question

For the Locus equation

$$\frac{\mathrm{d^2}u }{\mathrm{d} \theta^2} + u = - \frac{1}{h^2u^2}f\left(\frac{1}{u} \right )$$

How do I find the solution for $f(r)= \frac{-h^2}{r^3}$ and sketch the solution for some initial conditions?

Answer 1

$$u=\frac{1}{r}$$ $\therefore$ $$\frac{1}{u}=r$$

and so $\frac{-h^2}{r^3}=\frac{-h^2}{\frac{1}{u^3}}=-h^2u^3$

$\implies$ $$\frac{\mathrm{d^2}u }{\mathrm{d} \theta^2} + u = \frac{h^2u^3}{h^2u^2}$$

and so we have $$\frac{\mathrm{d^2}u }{\mathrm{d} \theta^2} =0$$

thus, $$u=A{\theta}+B$$

$\implies$ $$r=\frac{1}{A\theta+B}$$

hence, $$r=\frac{1}{1+\theta}$$

which gives us an inward spiral around the origin.

So we chose $B =1$ as otherwise it is undefined when $\theta=0$.

Initial Conditions:

$$A=1, B=1$$