Now that we know what to do when we have complex conjugate roots of the characteristic equation, we can try and work in an example. So, let's solve x double dot plus x dot, plus x equals zero with initial conditions, x of zero equals one and x dot of zero equals zero. We try our ansatzs as always. Try x equals e to the rt. We get the characteristic equation, r squared plus r, plus one equals zero. We use the quadratic formula here. So, r plus or minus equals negative b, negative one, plus or minus the square root of b squared one, minus 4ac minus four, over 2a two. We notice that those are complex numbers. So, this is minus 0.5 plus or minus i times root three divided by two. So, this is the case of complex conjugate roots of the characteristic equation. We know now how to find two real functions. So, let me write them down. We have, first let us write the general solution using our two real functions. So, we have x, x of t is equal to, we have a real exponential function in both terms, so e to the minus t over two. Then the complex, the imaginary part of r goes as the frequency in the cosine and the sine term. So, we have a constant A times cosine, root three over 2t, plus another constant B, times the sine root three over 2t. That's the general solution then in the case when we have complex conjugate roots of the characteristic equation. We have two free coefficients here A and B, and we have to use them to satisfy these two initial conditions. So unfortunately, we have to take the derivative of this. We going to have to use the product rule to do that. So, we're going to need x dot of t. So, let's do that relatively slowly. So, it's the derivative of the first times the second plus the first times the derivative of the second. So, the derivative of the first gives us minus 0.5e to the minus t over two times the second, which is A cosine root three over 2t plus B sine root three over 2t, plus the first times the derivative of the second. The derivative of the second is always going to, using the chain rule, is always going to bring out a root three over two. So, we have plus root three over two times e to the minus t over two. Then the derivative of cosine is minus sine. So we have minus A sine root three over 2t and the derivative of sine is cosine. So, plus B cosine root three over 2t. That's the derivative. It's unavoidable to have this long expression, but when we plug in x of zero and x dot of zero, things will simplify. So, let me write it up here, x of zero equals one. So, put in t equals zero, e to the zero is one, cosine of zero is one, sine of zero is zero. So, x of zero is just equal to A, and that's supposed to be equal to one. So, we get A equals one. How about x dot of zero? X dot of zero, we have minus 0.5 times e to the zero. So, minus 0.5. The sign will be zero and we'll end up with A. So minus 0.5 A. Then the second piece here, we have a plus root three over 2e to the zero is one, sine of zero is zero, cosine of zero is one. So, plus root three over 2B, and that's supposed to be equal to zero. So, we can use the second equation to solve for B. So root three B is equal to A, A is one, so root three B equals one, and then multiply by root three and divide by three. So, we get B is equal to root three divided by three. So, what is our solution here? Our solution for x is given by this expression, and then we've determined what A is and we've determined what B is. That's the solution then to the differential equation and the initial conditions. So, let me summarize what I've done. I'm solving x double dot, plus x dot, plus x equals zero, and looking for the solution that also satisfies these two initial conditions. When I try our ansatzs, x equals e to the rt, we end up with complex conjugate roots of the characteristic equation. Then we can use those roots to do the principle of superposition the first time to construct two real solutions. The two real solutions are e to the minus t over two, the minus 0.5 goes in the exponential, and cosine root three over 2t, that's from the imaginary part, and then sine root three over 2t from the imaginary part. So, we can just immediately write down the general solution of a differential equation with complex conjugate roots. Then we need to satisfy the two initial conditions. Unfortunately, we have to differentiate this, but then when we substitute in t equals zero, we get some relatively simple linear system to solve for A and B. Then we solve that, and then we have the solution that satisfies both the differential equation and the two initial conditions. I'm Jeff Chasnoff. Thanks for watching, and I'll see you in the next video.