# Binet's differential equation yields a circular path

Binet's equation is $$\frac{d^2u}{d\theta^2}+(1-\frac{\lambda}{a^2v^2})u=0$$
I need to show that for $\lambda=a^2v^2$ path will be a circle.
For $\lambda=a^2v^2$ I have $$\frac{d^2u}{d\theta^2}=0$$

Can you give me some hints what to do now?

• Convert from Polar Coordinates to Rectangular! You are being fooled by the form of the second equation in your post. Aug 26 '13 at 22:34
• What are Rectangular Coordinates? Cartesian maybe? I'm not sure if I know how to convert them. May you help me?
– gov
Aug 26 '13 at 22:57
• See here. Aug 27 '13 at 0:29
• Also, check this. Aug 27 '13 at 0:35
• I checked both but I am not sure what are you telling me to do. Maybe I should solve inverse problem?
– gov
Aug 27 '13 at 9:39