INSTRUCTOR: Find the charge on the capacitor in an RLC series circuit where L equals 1/5 henries, R equals 2/5 ohms, C equals 1/2 farads, and E of t equals 50 volts. Assume the initial charge in the capacitor is 0 coulombs and the initial current is 4 amps.

OK. Now, there's a lot going on here, but let's start out with just the known equation, and that is L times the second derivative of q plus R times the first derivative of q plus 1 divided by C, end fraction, times q equals E of t. Let's take all the values we know. We'll place them in here and see if we can reduce this to an easier solve-- more easily solved problem.

So this is equal to 1/5 times q-- or the second derivative of q-- plus 2/5 times the first derivative of q plus 2/5-- 1 over C. C is 1/2-- ah, plus 2q equals 50. Now I'm going to multiply every value here by 5 to eliminate fractions. So this will be the second derivative of q plus 2 times the first derivative of q plus 10q equals 250.

Now let's consider the complementary equation and the characteristic equation of that. So that would be lambda squared plus 2 lambda plus 10 equals 0. And, solving that, we can find that lambda is equal to negative 1 plus or minus 3i. Which means that our general solution is of the form e to the negative t, end exponent, open parenthesis, c sub 1 times cosine, open parenthesis, 3t, close parenthesis, plus c sub 2 times sine, open parenthesis, 3t, close parenthesis, close parenthesis. That's our general solution.

And based on the R of t that we have, 250 in this case-- based on that, we're going to make a preliminary guess that q sub p of t is equal to A. That's because 250 is a constant, so we're going to suppose that our particular solution is a constant. OK.

Now if we substitute that value, that function A our equation, the second derivative of A is-- of q sub p is 0. First derivative of that is also 0. So this gives us the equation 10 times A equals 250. So that A, our coefficient there, is 25. Or our function there is q sub p of t equals 25.

Now, summing these things together, we can have our general solution of q-- or our particular solution. It's the sum of our general solution and our guest function that we have with our parameter. So q of t equals e to the negative t, end exponent, open parenthesis, c1 times cosine, open parenthesis, 3t, close parenthesis, plus c sub 2 times sine, open parenthesis, 3t, close parenthesis, close parenthesis, plus 25.

Now, we know our initial charge. That has a value of q. So q of 0 equals 0. And the initial current, that is q prime at 0. That is equal to 4.

Now, finding the derivative is probably the hardest part here, just because it's your product rule. So we find our first derivative and we take our original function, evaluate them both at 0, and we come up with a system. c sub 1 plus 25 equals 0, and negative c sub 1 plus 3 times c sub 2 equals 4. Combine these together and we see that c1 is negative 25, and c2 has a value of negative 7.

So that our final solution here for charge is q of t equals e to the negative t, end exponent, open parenthesis, negative 25 times cosine, open parenthesis 3t, close parenthesis, minus 7 times sine, open parenthesis 3t, close parenthesis, close parenthesis, plus 25. And this is our equation for charge.