 As long as we avoid singular points, we can always find a series solution to the differential equation, and the series will be valid in the interval of convergence common to all of our coefficient functions. This works if F i of t are polynomials, since we can then compare coefficients. But what if the F i t are not polynomials? In that case, we make them into polynomials. So for example, let's say we have the second-order differential equation. First, we note that for 1 over 1 plus t squared, all values of t are regular. And so we'll get convergence for all values of t. Now if we multiply by 1 plus t squared, we obtain the differential equation, and now our coefficient functions are polynomials. We'll assume a series solution, y equals the sum from 0 to infinity, an t to power n. We'll need our first and second derivatives, and now we can substitute into our differential equation. So we need our second derivative. We need t squared times our second derivative. So we can multiply the terms of our series by t squared to get. And so there's our t squared plus our second derivative. And then finally, we need to add y itself. So now to add these series, we'll first want to make sure the powers on t are the same. So these are both t to the n, but this one is t to power n minus 2. So let's adjust it. Since we want the power on t to be equal to n, we'll let our exponent n minus 2 be equal to k. Now this substitution requires us to evaluate what n equals 2 is, what n is, and what n minus 1 is. So if n equals 2, k is equal to 0. And so that's the starting point of our series. n itself is equal to k plus 2, so that's this factor. The other factor, n minus 1, is equal to k plus 1, filling in the rest of our terms. Since it doesn't matter what we call the index variable, we'll call it n. And this gives us the series. So now all our powers of t are the same, but the series start in different places. So we'll have to adjust, and remember, you can always start later. So these two series start at n equals 0, we want to start them at n equals 2. So we'll adjust them. So in order to start this series at n equals 2, we have to split off the n equals 0 and n equals 1 term. So if n equals 0, the term of this series is going to be... And if n equals 1, our corresponding series term will be... For the other series, we want to start it at n equals 2. So we have to split off the n equals 0 term and the n equals 1 term. And so now these two series can be rewritten as a series that starts at n equals 2, plus some leftover bits. So we can rewrite our first series to start at 2. Our second series already starts at 2. And our third series can be rewritten to start at 2. So separating out our leftover bits and combining the series into a single series, all of which should be equal to 0. So now we need to find our coefficients. So to find the coefficients, note that because this has to be equal to 0 for all values of t, we require that our constant terms, a0 plus 2a2, equals 0. Our linear coefficient, a1 plus 6a3, must also be equal to 0. And in fact, all of the other coefficients must also be equal to 0. And if we rearrange this last equation we get, which tells us how to find later coefficients in terms of earlier coefficients. Our initial condition gives us that y0 is equal to 1. Since we assumed a series form of our solution, if t is equal to 0, y of 0 is going to be... And so a0 is equal to 1. Likewise, our initial condition gives y prime of 0 is equal to 1. So differentiating, we find that y prime of 0 is equal to a1. And so a1 is equal to 1. And so these give us the coefficients of the first two terms in our series and we can begin writing the series. Our recurrence relationship tells us how to compute an plus 2 from an. So if I want to find a2, well that will be when n is equal to 0, so substituting in n equals 0, finding our values, a2 is minus 1 half, which gives us the next term of our series. For a3 we'll have n equal to 1, substituting and simplifying, we find that a3 is going to be negative 1 sixth. And that gives us the next term in our series. a4 is going to be 1 twelfth, and so the next term in our series is going to be 1 twelfth t to the fourth. And we can continue to compute series terms for as long as we like.