Capacitor charge - LRC series circuit

[america8371]

In differential equations I was assigned a few circuit exercises. I'm having trouble understanding them, but I'm pretty sure that if can understand the first one then I can get the rest of them. The first exercise asks for time of capacitor charge and the others ask for things like capacitor maximum charge, steady state current, current/charge equations, etc.

Homework Statement

Find the charge on the capacitor in an LRC series circuit at t=0.01s when L=0.05h, R=2$$\Omega$$, C=0.01f, E(t)=0V, q(0)=5C, and i(0)=0A. Determine the first time at which the charge on the capacitor is equal to zero. Please give answer to four decimal places.

a. t = 0.5095s
b. t = 0.9595s
c. t = 0.0509s
d. t = 0.5959s
e. t = 0.9059s

Homework Equations

I'm not too sure but i think the correct diff. eq. to use is

L $$\frac{d^{2}q}{dt^{2}}$$+R $$\frac{dq}{dt}$$+ $$\frac{1}{C}$$ q = e(t)

The Attempt at a Solution

I attempted to solve it by method of undetermined coefficients.

q''+40q'+2000q=0

r$$^{2}$$+40r+2000=0 r=-20$$\pm$$40i

q$$_{p}$$(t) = Ae$$^{-20t}$$cos 40t + Be$$^{-20t}$$sin 40t

etc.

It feels like I'm going in the wrong direction because it seem like there's more to this exercise than just solving the diff. eq.

 

[HallsofIvy]

Well, yes, there is more! But you are going in exactly the right direction. You just haven't yet answered the question! You are told that q(0)=5C, and i(0)= q'(0)= 0A. That let's you find A and B. And then you need to determine the value of t such that q^p(t)= 0.