So, in this video let's solve an inhomogeneous differential equation, and I'll use the example where the inhomogeneous term is an exponential function. We'll do the full solution in this video where we also satisfy the two initial conditions. Okay, so remember it's a three-step process. So, the first step is to solve the homogeneous equation. We try our usual ansatz. Try x equals e to the rt. So the right hand side now is 0 in the homogeneous equation. So we get the characteristic equation r squared minus 3r minus 4 equals 0. We see if it's factorable, so 4 is 4 times 1. So we can have a minus 4 and a plus 1. So r minus 4r plus 1 equals 0, is the factor. Then we have two roots. So, we have the homogeneous solution then is equal to, a constant times e to the 4 t, plus another constant times e to the minus t. So we found the homogeneous solution. In step two we have to find the particular solution. So how do we find the particular solution? We're looking for any solution of the differential equation so that when you substitute x into the left-hand side, we get 3e to the 2t. So what should we try to substitute for x? We would like to convert the differential equation into an algebraic equation. So we would like to match the e to the 2t term on the right-hand side. So if we substitute in x equal to a constant times e to the 2t, we can get the e to the 2t term to cancel. So in this second step, we're looking for the particular solution. So we can try, x equals some unknown constant times e to the 2t. So A now is our unknown parameter that we're going to try and solve by substituting this into the differential equation. So when we start taking derivatives, we're always going to have an e to the 2t term on both sides of the equation, and we can cancel the e to the 2t term. So let's substitute n. We'll cancel the e to the 2t term. So x double dot will bring down 2 times 2. So x double dot will give us a 4A minus 3x dot will be minus 3A times 2. So minus 6A minus 4x will be minus 4A and that will be equal to 3 on the right-hand side. So I've canceled the e to the 2t term converted to an algebraic equation. So 4A minus 4A is 0. So minus 6A equals 3 since A is equal to minus one-half. Okay. So we've got the particular solution. So we've got the homogeneous solution and the particular solution. So then we're on to step three, which is to write x as a sum of the homogeneous solution plus the particular solution. Okay? This x is the particular solution. So let me write xp here. Okay. So we put it together. So, this is equal to the homogeneous solution. So C1e to the 4t plus C2e to the minus t plus the particular solution, so minus one-half times e to the 2t. Okay, and then we satisfy the two initial conditions. So, using this form for x of t the homogeneous solution plus the particular solution, we need x of 0 equals 1. So, x of 0 equals 1. So x of 0 we'll get C1 plus C2 minus one-half equals 1. X of 0 equals 1. X dot of 0 equals 0. If we take the derivative of this and put in t equals 0. We'll get 4C1 minus C2 and then the derivative will be minus 1, and that's supposed to be equal to 0. x dot of 0 equals 0. So this is a two linear equations. So let me write that here in a different color so you can see. C1 plus C2 equals three-halves, and second one will be 4C1 minus C2 equals 1. Okay, so that's our two linear equations for the two unknowns C1 and C2. What are the solutions here? We can add them, right? And then the C2 will go away. We'll get 5C1 is equal to five-halves, right? Adding them 5C1 equals three-halves plus two-halves is five-halves. So C1 equals one-half, right? So by adding them we get C1 equals one-half. Then we get C2, so one-half plus 1 is equal to three-halves. So C2 is equal to 1. Okay, so we get our final solution. We can write it in a nice form. So let me write it down here. On the bottom here. Very small light board. So, we put it together. We get x of t, I'm going to put it in a dominant terms first. So the dominant term here is C1e to the 4t, so that's one-half. e to the 4t, the next term is minus one-half e to the 2t, and the least dominant term is this decaying term. So C2 1 times e to the minus t, plus e to the minus t. Okay and that's our solution then. This solves both the differential equation and the initial conditions. We could write it in a slightly nicer form. We can factor out the one-half, e to the 4t, and then we can have a 1. Then we have a one-half here e to the 2t. So that will be a minus e to the minus 2t. Then the factoring out a one-half e to the 4t will have a plus 2e to the minus 5t. Okay. That's a little bit cleaner form because you see that the solution grows like one-half e to the 4t, and then everything else will eventually decay to 0. Okay, so let me summarize. We're solving now a second-order linear inhomogeneous equation, but also with constant coefficients. We're using a three-step process. We found the homogeneous solution here. A new step is to find a particular solution. The onslaughts we use to find a particular solution depends on the form of the inhomogeneous term. If the inhomogeneous term is an exponential function, we used an exponential function onslaughts. We match the e to the 2t, but we have a 3 parameter here. When we substitute into the differential equation we can determine the free parameter. Here it's A equals minus one-half. In the third step, we combine both the, we add the homogeneous and the particular solution. We have two free constants. So, we use those to free constants to satisfy the two initial conditions. We can determine C1 and C2, right? A solution and then if we want to clean it up, we can collect the dominant term, factor out the dominant term. Okay. In the next videos we'll see how to find particular solutions for other types of right-hand sides. I'm Jeff Jasnoff, thanks for watching, and I'll see you in the next video.