# First Order DE problem (electric circuit)

1. May 11, 2008

### forty

(dq/dt) + (5/(20+t))q = 20/(20+t) ... t=0 , q=1

(a) Find the charge on the capacitor at any time.

This is linear so i found the integrating factor which is (20+t)^5

and solved q(20+t)^5 = 20 integral ((20+t)^4) dt

and got q(20+t)^5 = 4((20+t)^5) + C

my C value i got was -9600000

solving for q gave me q = 4 - ((9600000)/((20+t)^5))

(b) Transient and steady state solution

Transient = - ((9600000)/((20+t)^5))

(c) Find the current in the circuit at any time, hence the maximum current in the circuit.

current = i = dq/dt

I differentiated and got dq/dt = 48000000((20+t)^-6)

**My problems

Firstly does everything follow, is my logic correct?
Secondly for the maximum current that occurs at t = 0 (presuming my equations are correct) is there anyway of showing working for that, other then stating its at t = 0? (t = 0, i = .75)

2. May 11, 2008

### HallsofIvy

Staff Emeritus
Yes, everything you've done appears to be correct. As for showing that the maximum current occurs at t= 0, it is sufficient to observe that 48000000(20+t)-6 is a decreasing function. Simply observing that as t increases, the denominator increases, therefore the function decreases is enough- or if you could show that the derivative of i is never 0- therefore has a maximum at an endpoint, t= 0.

3. May 11, 2008

### forty

OK so just observation is the way to really go, too easy thanks alot!