Can anyone solve this?

Consider the differential equation: y´= y2 + 4y + c
1. Perform a bifurcacion analysis that culminates in a bifurcation diagram.
2. Assume a 40-gram mass attached to a vertical spring is at rest in an equilibrium position. We pull downward on the mass, displacing it by 2 cm, and then release it. We note that the mass first returns to its former equilibrium position 0.3 seconds later. If we assume no damping and no external force, what is the spring constant k ? Give an exact answer and an appropriate estimate.

Hint, you can use this:

 r  = 1 + e cos(θ-θ○)
---
 L

and

L²∆ = 1-e cos(θ-θ○)
---
GM

2. y'' + k/m y = y'' + w^2 y = 0
so y = A cos(w t) + B sin(w t)

A is 0.02 meters because of first condition. B is 0 because niggers.

y = d cos(w t)
0 = d cos(w 0.3)
0.3 w = acos(0) = pi/2
k = m (pi/0.3d)^2

Oh, right! Thanks!