 So as a general rule, most differential equations are unsolvable. At least in the sense of being able to write down a recognizable function that will solve them. But remember, your inability to solve a problem does not make the problem go away. We still need to find some method of representing the solution, and so for that, we turn to the Taylor series. Remember that a function f, which has derivatives of all orders at x equals x0, has a Taylor series centered around x equals x0, and this Taylor series gives the correct value for f of x0. And if we're lucky, this series will give us the correct value of f of x for other values of x as well. We define the following. A function f is analytic at x equal to x0 if the Taylor series centered at x equals x0 converges to f of x for some open interval containing x0. The short version of analytic is that within the interval of convergence, an analytic function has a Taylor series that converges to the function itself. And in addition, there are several more useful properties. As long as we stay inside the intervals of convergence, the series converges absolutely, and more importantly, the series can be integrated or differentiated term by term. And in either case, the interval of convergence remains the same with the possible exception of gaining or losing an endpoint. And this suggests a powerful method of solving differential equations. First, assume a power series solution, and then find the power series. For example, let's try and solve this differential equation where y of 0 is equal to 0. So we'll assume a power series solution, y equals the sum from n equals 0 to infinity of an x to power n. We'll also assume that this is analytic, although that's a bit of wishful thinking at this point. Provided it is analytic, then we can find the derivative by differentiating term-wise, and so our derivative will be, and so our differential equation gives us. Now in order to proceed, we'll need to reindex our terms. So again, what we want is our exponent on the variable to be the same, so we want n minus 1 to be equal to k, and so n is equal to k plus 1, and our starting value n equals 1 corresponds to k equals 0, and so our series becomes, and since the name of the index variable doesn't matter, we'll call it n and get our series, and now our two series have the same power on x. They both start at n equals 0, and so we can combine the coefficients, and that gives us our differential equation expressed as a power series. To find our coefficients, we'll assemble our equation dy dx plus y equal to 1, so we know our power series for dy dx plus y, and it may be helpful to write out a few terms of our power series. If n equals 0, our corresponding term will be, if n equals 1 we have, n equals 2 gives us, and so on. In order to find these coefficients, we remember that we have to have equality, since we know y of 0 is equal to 0, and by assumption y is our power series, we can compare the two. If x equals 0, our power series becomes, and y is just 0, and so a0 is equal to 0. Since a0 is equal to 0, and our left-hand side and the right-hand side have to be the same, that means our constant terms have to be the same. On the right, we have 1. On the left, our constant term is a1 plus a0, and so these must be equal. Solving tells us that a1 is equal to... Again, our left and right-hand sides have to be equal. On the left, our coefficient of x is 2a2 plus a1. On the right, our coefficient of x is 0, and so this gives us the equation, which we can solve for a2. Again, because the left and right-hand sides have to be equal, our coefficient of x squared on the left has to be the same as the coefficient of x squared on the right, and so that gives us the equation, which we can then solve for a3. And for purposes of later recognitions, it's sometimes useful not to do all the arithmetic, so we'll leave a3 in this form. But wait, there's more. The coefficient of x to the third on the left is... Well, that's not written down, but we can compute it from the form of the series coefficients, and so that will be... and on the right, because there's no x to the third term, that coefficient is going to be 0, and so we have the equation, which we can solve, and this gives us the first few terms of our series. In some cases, it may be possible to recognize a power series as a familiar function. In this case, note that e to the power x has power series, which means that e to the power minus x will have the power series, and with a little bit of algebra, we see that this differential equation, which has this solution as a power series, has the same power series as 1 minus e to the power minus x, and so our solution is y equals 1 minus e to minus x. However, as a general rule, getting a closed form expression for a series, in other words, being able to identify a power series as a familiar function, is usually impossible.