So, we're trying to solve the homogeneous, second order differential equation with constant coefficients. There are three cases we have to consider. So, the first case is when this quadratic polynomial, what we call the characteristic equation, has distinct real roots. So, let's work an example, so you can see how to get the solution. Let's consider the constant coefficient equation, x double dot plus 5x dot plus 6x equals 0, and we have two initial conditions: x of zero equals two and x dot of zero equals three. So, we solved a constant coefficient equation by an ansatz. So, we try x equals e to the rt. We substitute into the differential equation, and we're going to cancel e to the rt, so we end up with the quadratic equation r squared plus 5r plus 6 equals 0. You can use the quadratic formula, but here, actually, it factors. Three times two is 6 and three plus two is five. So, this becomes r plus 2, r plus 3 equals 0. We have the roots r1 equals minus two and r2 equals minus three. So, this is the case of two real roots of this characteristic equation. So, we can use the principle of superposition. We've found two solutions. x equals e to the minus 2t and x equals e to the minus 3t, and then we can super impose those solutions. So, we have our x of t then is equal to c1 e to the minus 2t plus c2 times e to the minus 3t. I claim that this is the general solution to this differential equation, because the ranskian of these two functions is not zero. We don't need to show that here, but you can take my word for that. So, we have to satisfy the condition x of zero equals two, but also the derivative, so we can compute the derivative here. So x dot of t is minus 2c1e e to the minus 2t minus 3c2e to the minus 3t. Satisfy the initial conditions. So, we plug in t equals zero here. So, the first equation will give us c1 plus c2, and that will be equal to x of 0, which is 2. The second equation, x dot of 0 equals 3, will give us minus 2c1 minus 3c2 equals x dot of 0, which is 3. So, we have then two linear equations and two unknowns. The unknowns are c1 and c2. There are many methods to solve this. Maybe the easiest way is to eliminate c2. So, we can take this first equation and multiply it by three, and then add them, and then c2 will be eliminated. We get 3c1 minus 2c1, which gives us c1, and then we get 6 plus 3, which will give us 9. So, c1 is 9. c2 then is 9 plus 1 equals 2, 9 plus negative 7 equals 2, so c2 is negative 7. Putting it all together. Then we have our x of t. So, let me write it here. We get x of t equals c1, 9 times e to the minus 2t plus c2 minus 7 times e to the minus 3t. That's the solution of the differential equation. That also satisfies the two initial conditions. If we wanted to write that in a nice form, we would factor out the slower decaying solution. So, we'd have 9e to the minus 2t, and then we would have a 1 minus 7 over 9 times e to the minus t, that's how a mathematician would write such a solution. So, you can see that the 7/9th e to the minus t decays, first compared to the one, and then the whole thing decays like 9 times e to the minus 2t. So, let me summarize. We're solving the second order, a homogeneous differential equation with constant coefficients, trying to satisfy two initial conditions, x of zero and x dot of zero are given. We use our ansatz, x equals e to the rt, we get our characteristic equation, r squared plus 5r plus 6 equals 0. In this case, the factors, and we have two distinct real roots. r1 is minus 2, r2 is minus 3. We use the principle of superposition to write the general solution. We need the time derivative of that, and then we satisfy the two initial conditions, that gives us two linear equations for the two unknowns, c1 and c2, and then we can find the specific solution that satisfies both the differential equation and the initial conditions. In the next two videos, we'll consider the other two cases. I'm Jeff Jasnoff, thanks for watching, and I'll see you in the next video.