Finding the Second Constant in an XRLC Circuit Current Response Problem

[bodomizer]

Homework Statement

FIND i(t).

[/B]

Homework Equations

VL = L*(di/dt) (Equation 1)
IC = C*(dV/dt) (Equation 2)

I(t) = e^-9*t*(A*cos(4.359*t) + B*sen(4.359*t)) A (Equation 3)

The Attempt at a Solution

I can get the exact value of the constant B as shown in the answer if I use -6 after the equal sign of the integral. The problem is, I don't understand WHY it should be -6. I solved everything else correctly, but I don't understand completely how to find the second constant "B".

PS: I solved the integral on Wolfram Alpha.

 

[phyzguy]

I would look at it this way. Immediately after opening the switch, do a KVL around the loop. The voltage across the capacitor is 6V, because that's what it charged up to while the switch was closed. 1A of current is circulating around the loop, because that's how much current was flowing when the switch was closed. So each of the resistors has 1A flowing through it, so the 6 ohm resistor is dropping 6 volts, and the 3 ohm resistor is dropping 3V. So summing up the voltages around the loop gives:
$$6 + L \frac{di}{dt}(t=0) + 3 -6 = 0$$
$$L \frac{di}{dt}(t=0) = -3$$
$$0.5 (-9 + 4.359 \times B) = -3$$
$$B = \frac{3}{4.359} = 0.6882$$

 

[bodomizer]

@phyzguy Your solution seems correct, but shouldn't I find the initial conditions before opening the switch? That's how it worked for all the other exercises: I find the ODE on t>0 and solve it, and then the constants from the initial conditions on t = 0-

 

[phyzguy]

@phyzguy Your solution seems correct, but shouldn't I find the initial conditions before opening the switch? That's how it worked for all the other exercises: I find the ODE on t>0 and solve it, and then the constants from the initial conditions on t = 0-

I don't see how. Before you open the switch, there is no voltage dropped across the inductor, since di/dt = 0, and there is no voltage dropped across the 3 ohm resistor, because there is no current flowing through it. It is only after the switch is opened that the initial conditions are established. For example, if I change the 3 ohm resistor to 3000 ohms, all of the initial conditions when the switch is closed are the same, but the initial conditions after the switch is opened are very different. I would say that the constants are found from the initial conditions at t=0+, not t=0-.