1. SERIES SOLUTIONS TO DIFFERENTIAL EQUATIONS AP Calculus BC

2. OMG! WHY????? Because many differential equations cannot be solved in terms of combinations of simple functions. Even y ʹʹ – 2xy ʹ + y = 0 can’t be solved this way, but equations like this arise in the “real world”, such as in quantum mechanics. So our solution will be of the form The method is to substitute this expression into the differential equation and determine the values of the coefficients. It is similar to the method of undetermined coefficients.

3. EXAMPLE 1 Solve the equation y ʹʹ + y = 0 Yes, we could use easier techniques to do this, but we’re going to use power series. Starting out easy (sorta)! We assume there is a solution of the form Differentiate term by term:

4. EXAMPLE 1 (CONTINUED) It’s easier to compare y and y ʹʹ if we rewrite y ʹʹ like this: Now substitute into the differential equation: Combine:

5. EXAMPLE 1 (CONTINUED) Therefore  Solving for c n+2  This is called a recursion relation. If we know c 0 and c 1, we can determine the rest of the coefficients by putting in n = 0, 1, 2, 3, … Try it. See a pattern?

6. EXAMPLE 1 (CONTINUED) For the even coefficients: For the odd coefficients: Substitute back into the original equation, which was:

7. EXAMPLE 1 (FINAL) So  And finally  Normally, we wouldn’t be able to express these as functions, but in this case 

8. EXAMPLE 2 Solve y ʹʹ – 2xy ʹ + y = 0 As before, we assume there is a solution of the form So and Substitute into the differential equation:

9. EXAMPLE 2 (CONTINUED) So And Therefore  Solve for c n+2 

10. EXAMPLE 2 (CONTINUED) Put in n = 0, 1, 2, 3, … and look for a pattern. The even coefficients are given by The odd coefficients are given by So the solution is

11. EXAMPLE 3 – INITIAL VALUE PROBLEM Solve y ʹʹ – 2xy ʹ + y = 0 where y(0) = 0 and y ʹ (0) = 1 General solution is Substituting in x = 0 leaves y = c 0 So c 0 = 0, and the even terms all drop out! y ʹ = (c 1 x) ʹ = c 1, so c 1 = 1 Solution is