CHAPTER 08.01: PRIMER ON ORDINARY DIFFERENTIAL EQUATIONS: Exact Solution of 1st Order ODE

In this one, we will look at an example of an ordinary differential equation. We'll look at the first-order ordinary differential equation, and we're trying to find an exact solution to this ordinary differential equation. So let's take an example here, 3, first derivative of y with respect to x, plus 2 y is equal to e to the power minus x, and you are given the initial condition of y at 0 to be 5. So how do we go about solving this? Now, there are several techniques available. In this segment, I'm just going to concentrate on showing you the classical solution technique, which is just finding the homogenous part and particular part, and then applying the initial conditions to find the solution. Now, the homogenous part can be found by writing this equation, this differential equation in using the operator here. So because D stands for the first derivative operator, so that's how you're going to do, so you are using the operator of 3 D, plus 2, on y is equal to e to the power minus x, y of 0 equal to 5. So if you are trying to see what the homogenous part of the solution will be, that will be based on what is the operator part, that will give you the characteristic equation for that. So the characteristic equation is 3 m, plus 2 equal to 0, so that's how you get the characteristic equation, by looking at what the operator is. So you're getting 3 m, plus 2 equal to 0, so m is equal to -2/3. So since this is the root of the characteristic equation, then in that case, the homogenous part of the solution will be k1 e to the power -2/3 x, that's what will be what the homogenous part of the solution will look like. Now, in order to find the particular part, so how do we find out the particular part of the solution? Well, the particular part depends on the right-hand side, or your forcing function, as you may call it. The right-hand side is e to the power minus x. So what you've got to do is in order to be able to choose the particular part will be to choose the form of the right-hand side, which is e to the power minus x, and all its possible derivatives, and all the possible derivatives of e to the power minus x are e to the power minus x itself, so that means that the only term which you will have in the particular part will be just A e to the power minus x, because if you take all possible derivatives, you're going to get simply another constant times e to the power minus x, another constant times e to the power minus x, so that's going to all bundle up into one expression which will look like that. So the particular part is basically dependent on the form of the right-hand side, and all its possible derivatives. So since we have the particular part, now we're going to substitute this particular part in our differential equation, we have 3 dyP by dx, plus 2 yP is equal to e to the power minus x. So now I'm going to substitute this particular part in there. It will be A e to the power minus x, plus 2 A e to the power minus x equal to e to the power minus x. And this gives you -3 A e to the power minus x, that's the derivative of that, it will be minus A e to the power minus x, plus 2 A e to the power minus x equal to e to the power minus x. That gives me minus A e to the power minus x equal to e to the power minus x, so the only way this equation can be resolved is by choosing A equal to -1. So the particular part of the solution in this case will be A e to the power minus x, which will be simply minus e to the power minus x, that's what the particular part of the solution will look like. So at this stage, what you do is, you already found out the form of the homogenous part, you just calculated what the particular part of the solution will be, and you will go ahead and add the two. Say y is equal to yH plus yP, and that's k1 e to the power -2/3 x, minus e to the power minus x. So the question arises that how do I find out what the value of k1 is? How do I find out what the value of k1 is? And the way you do that is simply substituting the initial condition, y of 0 is equal to 5, that's k1 e to the power -2/3 0, minus e to the power -0, so you get 5 is equal to k1 minus 1, you get k1 equal to 6. So that's how you are able to find out what the value of the coefficient of the homogenous part of the solution is. So once you have found k1 equal to 6, you substitute it back in into this equation, you get y is equal to 6 e to the power -2/3 x, minus e to the power minus x, and that's the solution to your first-order ordinary differential equation which you just got. As part of your homework, what I would like you to do is, what if the forcing function or the right-hand side which you had, was e to the . . . was 2 e to the power -2/3 x? So if you would change the problem to the right-hand side being 2 e to the power -2/3 x, what you are seeing is that this forcing function which you have here is same as the homogenous solution. So when you are doing that, what you have to do is you have to choose the next independent solution for your particular part, so I'm going to give you a hint. Your particular part will look like this, A x e to the power -2/3 x, because if you choose the form of the right-hand side itself and its derivatives to be the particular part of the solution, you're going to get 0 equal to a nonzero number, so you have to choose the next independent solution, which will be A times x e to the power -2/3 x to be able to solve that problem. So that's your homework. And that's the end of this segment.