I'm studying for my final undergrad assessment, so I thought a good way to study would be to make "advanced" versions of problems I'm already familiar with. Here's one I came up with but I'm not quite sure I'm on the right track exactly.

**1. Homework Statement**

A rope, of mass [tex]M[/tex] and length [tex]L[/tex] (and it's uniform density [tex]\lambda[/tex], for simplicity) is suspended over an idealized pulley (no masses attached to the rope for now). It's released from rest at time [tex]t = 0[/tex] with the length of one side (The right side, for the heck of it) being length [tex]y[/tex]

_{0}. I want to find the equation of motion of the string.

**2. Homework Equations**

I like using Energy methods, so I'll go with that here:

[tex]E = T + V[/tex]

[tex]T = 1/2 m \dot{y}^2[/tex]

[tex]V = m g y_{cm}[/tex]

**3. The Attempt at a Solution**

Find the total Kinetic Energy of the system:

[tex]T = T_{right} + T_{left} = 1/2y\lambda\dot{y}^2 + 1/2(L-y)\lambda\dot{y}^2 = 1/2M\dot{y}^2[/tex]

And the total potential:

[tex]V = V_{right} + V_{left} = -\frac{\lambda g}{2} y^2 - \frac{\lambda g}{2}(L-y)^2 = -\frac{\lambda g}{2}(L^2 - 2Ly + 2y^2)[/tex]

So the total energy is constant and has the value

[tex]E = T + V = 1/2M\dot{y}^2 - \frac{\lambda g}{2}(L^2 - 2Ly + 2y^2)[/tex]

So lets plug in our initial conditions to find the equation of motion at any time!

At [tex]t=0, y=y_0, \dot{y}=0[/tex]:

[tex]-\frac{\lambda g}{2}(L^2 - 2Ly_0 + 2y_0^2) = 1/2M\dot{y}^2 - \frac{\lambda g}{2}(L^2 - 2Ly + 2y^2)[/tex]

Which simplifies to:

[tex]2g(Ly_0-y_0^2) = L\dot{y}^2 + 2g(Ly-y^2)[/tex]

This is a differential equation of the form [tex]\dot{y}^2 + \alpha y^2 + \beta y = C[/tex].

I have no idea how to solve this final differential equation. Is it possible that if you choose some clever variable to substitue in you can turn it into an easily solvable linear equation? Or did I mess up somewhere in my Energy equations?

Thanks in advance!