 Suppose we have an nth order linear differential equation. Now, since our ability to use a power series requires that we actually have an analytic function that converges to the power series, we do have to make the following restrictions. We say that t equals t0 is an ordinary point if our coefficient functions are analytic at t equal t0 for all i and a singular point otherwise. We don't like singular points. But as long as we're near an ordinary point, a convergent power series solution always exists and it will converge on the common interval of all of our coefficient functions. For example, let's see if we can find a power series solution to this differential equation. Since the coefficients 1 and 5 are analytic everywhere, we assume y equals some power series. Our derivative is, and we'll re-endix this so it's in standard form and our series start in the same place. Our differential equation also requires 5y, which will be, and so dy plus 5y is going to be. To find the particular solution, we use the fact that we know y of 0 is equal to 10. So under our assumption that y is this series, then y of 0 is a0, equals means replaceable, so y0 equals 10, and so that tells me a0 equals 10. Now since we have to solve the differential equation, we know that dy dt plus 5y, which has this series form, must be equal to 0. But the only way for that to happen is that all of our coefficients have to be 0. And so we know that n plus 1 an plus 1 plus 5 an must be 0, and we have this relationship among the coefficients. Since we want to know what the coefficients are, a useful strategy is to solve for the later coefficients in terms of the earlier coefficients. So I want to solve for an plus 1 in terms of an. So solving gives us, and so now I know how to find the coefficients. Now once we've identified that we're going to find a series solution and have some formula that tells us the coefficients of our series, we have a solution. So this is our solution, but let's actually find a couple of the coefficients so we can get some feel for what this solution looks like. So we already know a0, and we know how to find an plus 1 from an. So if n equals 0, our formula gives us substituting our value for a0, a1 equals minus 50. But now we know a1, so if n equals 1, our formula gives us a2 is 125. But wait, now we know a2, so our formula gives us a3 equals minus 625 thirds. And so on, we can continue to calculate coefficients as far as we wanted. Now a good rule in math and in life is to make sure that new methods give you the same solutions as old methods. So this is actually a differential equation we should be able to solve, and in this case the solution is in fact y equals 10e to power minus 5t. And so y equals 10e to power minus 5t, and we can write the power series for this function, and we can find the first few coefficients. And we see that the coefficients of the power series which we know is the solution are the same as the coefficients of the power series solution. So the fact that we got the same solution using the power series that we would have gotten if we had used a method we already knew gives us some confidence that this power series solution actually works in cases where we don't already know the solution. So let's take a look at that next.