In this video, we'll start to apply the Laplace transform to a constant coefficient, second-order ODE. So, we need to transform the differential equation to get an algebraic equation. So, let's see how that works. We start with a second-order differential equations. So, we'll have ax double dot plus bx dot, plus cx, and there'll be an inhomogeneous term and we'll call that g. So, that will be a function of time. We take the Laplace transform of this equation, meaning taking the Laplace transform of both sides and we use the linearity of the transform. So, we can apply the transform to each term and pull the constants out. So, we will end up with a, times the Laplace transform of x double dot, which is a function of time, plus b times the Laplace transform of x dot, plus c times the Laplace transform of x, is equal to the Laplace transform of g, the right-hand side. Okay. So, we've Laplace transform the differential equation. The Laplace transform of x of t, is what we call capital X of s. The Laplace transform of g of t, is what we will call capital G of s. But the stumbling block here is we need to find the Laplace transform of the derivatives of x. To do that, we need to make use of the definition of the Laplace transform. So, let's look at the Laplace transform of x dot. So, the Laplace transform of x dot, which is a function of time, is going to be an integral from zero to infinity of e to the minus st, times x dot of t dt, and we need to do this integral, but we don't need to actually solve the integral. Well we need to do is express this integral in terms of a Laplace transform of x. If we can get every term here in terms of the Laplace transform of x, then we can solve for that capital X of s. So, how do you do that? Well, we have this pesky derivative on x. So, we need to pull that derivative over, which is what you do when you do an integration by parts. So, we have to do an integration by parts. We have a u and a DV piece. So, we can let U equals e to the minus st, and then dv is equal to x dot of t dt. We can differentiate u, so du, is minus Se to the minus st. We can integrate dv, which gives us V, which is just x of t. Okay. Then, we use this integration by parts. Let me put it here. So, we get L, the Laplace transform of x dot of t, then it is going to be equal to uv. So, e to the minus st, x of t, as t goes from zero to infinity, minus vdu. So, minus the integral of vdu, will be plus. So, will be plus the integral from zero to infinity, vdu has an s term and then we have an e to the minus st, x of t dt. Okay. So, what do we have here? We have the boundary term. We're always going to assume that, as t goes to infinity, e to the minus st, times x of t goes to zero. So, s is large enough so that will hold always. Then, we have the lower limit which will be minus x of zero. So, this is equal to minus x, at t equals zero. That will be the initial condition of x. So, the Laplace transform technique will need to solve an initial value problem and then this last term, this integral is just capital X of s. So, less just the Laplace transform of x. So, we have s plus our capital X of s. Okay. So, the Laplace transform of x dot by an integration by parts, we've gotten s times the Laplace transform of x, minus the initial value of x. In a similar way, we can find the Laplace transform of x double dot. The technique here, is that you do an integration by parts until you get to the Laplace, so that you get to the Laplace transform of x dot, and then you use the previous result. I don't need to repeat that algebra here but what you end up getting is s squared, times the Laplace transform of x, which is capital X of s, minus s times the initial value of x, x of zero minus x dot of zero. So, you need the initial value for the derivative of x. Okay. So, at this point now, we've Laplace transform, the x double dot and we have Laplace transformed x dot. So, now we're at the differential equation level. So, what does the differential equation become in s space. a, times the Laplace transform of x double dot. So, we end up with a, times s squared, capital X minus s x of zero. So, x of zero, we can assume to be x naught, so, minus sx naught, the initial value of x, minus x dot, which will be minus u naught. Okay. So, x of zero is x naught, x dot of zero is u naught. That's the first term, a Laplace transform of x double dot, plus b times the Laplace transform of x dot, which is s times capital x, minus x sub zero which is our x naught, and then plus c, times the Laplace transform of x and that's equal to the right-hand side which is our capital G of s. This is the equation in s space, after taking the Laplace transform of the differential equation in t space. It's a linear equation in capital X and you can easily solve it, right? You can isolate the capital X term on the left, everything else on the right and then divide through by the coefficient of capital X and solve this for capital X of s. Very simple because this is a linear equation. Okay. The other point that I should make here again, is that this equation contains the initial values for x, x of t, x of zero and x dot of zero, here x naught and u naught. So, the Laplace transform technique, takes the differential equation for second-order plus two initial conditions and gives you an algebraic equation for the Laplace transform of x of t which you can solve. Okay, let me review again. We start with a differential equation in t space, constant coefficient second-order with an inhomogeneous term. We take the Laplace transform of that equation. That means we should be able to take the Laplace transform of this inhomogeneous term. We can use the definition of the Laplace transform to figure out what is the Laplace transform of the derivatives in terms of the Laplace transform of the function itself and the initial conditions of the problem. When we do that, then we end up with an algebraic equation for capital X of s, which is very simple to solve because it's a linear equation. The final step, will be then taking the inverse Laplace Transform of capital X of s and we'll use a table to do that. Okay, as one more point I should make, nobody ever remembers these formulas or wants to recalculate it each time. So, I've put them in our table, which you can get from the lecture notes. Let's just have a quick look at that table before I say goodbye. So, in the table, we have on line 15 and 16, line 15 is the Laplace transform of the derivative of x and line 16 is the Laplace transform of the second derivative of x. Okay. In this section, we will make a lot of use of this table, because there's no point in repeating all this algebra when you have to solve a differential equation. Am Jeffrey Chasnov, thanks for watching and I'll see you in the next video.