In this exercise we consider finding the power series...

In this exercise we consider finding the power series solution of the second order linear initial value problem 1y'' - xy' + 4y = 0 subject to the initial condition y(0) = 1, y'(0) = 1. Since the equation has an ordinary point at x = 0, it has a power series solution in the form y = ?_{n=0}^{?} c_n x^n (1) Insert the formal power series into the differential equation, we derive an equation [ ] c_2 + [ ] c_0 + ?_{n=1}^{?} (([ ]) c_n + [ ] c_{n+2}) x^n = 0 So we have c_2 = [ ] c_0 and a recurrence relation c_{n+2} = [ ] c_n for n = 1, 2, ... (2) We can find a solution by letting c_0 = 1 and c_1 = 0: y_1 = 1 + [ ] x^2 + [ ] x^4 + [ ] x^6 (3) We can find another solution by letting c_0 = 0 and c_1 = 1: y_2 = x + [ ] x^3 + [ ] x^5 + [ ] x^7 ... (4) The general solution is given by y = Ay_1 + By_2 with arbitrary constants A, B. Note that y(0) = Ay_1(0) + By_2(0) = A and y'(0) = Ay_1'(0) + By_2'(0) = B Now, find A and B for the initial value problem A = [ ] and B = [ ] .

Transcript

00:01 All right, we want to use power series to find the solution to this.

00:07 I already worked it out once, so now i know what's going to happen, so hopefully i won't get you confused.

00:14 Okay, so here's the power series, so we need its derivative.

00:17 Oops, that was n equals zero.

00:19 So the derivative starts at 1.

00:22 Cn is a constant, so nx to the n minus 1.

00:27 And the second derivative starts at n equals 2, cn, n, or constant.

00:34 So n minus 1, x to the n minus 2.

00:39 So now we put it in and we get n equals 2 to infinity, cn, n minus 1, x to the n minus 2, minus x to the n minus 2, minus x times the first derivative, n equals 1 to infinity, cnnn x to the n minus 1, plus 4 times the function, n equals 0 to infinity, cn, x to the n.

01:06 That equals 0.

01:10 Okay, next i'm going to move the x and the four inside.

01:14 So n equals 2 to infinity, cn in n minus 1, x to the n minus 2, minus n equals 1 to infinity, cn, so when you multiply x times x to the n minus 1, you get x to the n for cn, x to the n.

01:40 Okay, now what you have to do is you make sure that inside, each of these, the x has the same exponents.

01:47 We have in here, we have in here, and we have n minus 2 here.

01:51 So we need this n minus 2 to be n.

01:56 But so i don't get confused, i'm temporarily going to call it m.

01:59 I'm going to let m equal n minus 2.

02:02 So m plus 2 is n.

02:04 So i'm just working on this part for right now.

02:09 So i have m plus 2 equals 2 to infinity, cm plus 2, 2, m plus 2, m plus 2 minus 1, and x to the m plus 2 minus 2.

02:32 Okay, i'm working on it still.

02:35 So that's m equals 0 to infinity, cm plus 2, m plus 2, m plus 2, m plus 1, x to the m.

02:50 All right, so now i'm going to put it back into the equation, only now i'm just going to change his name back to n, so they all match up.

02:58 N equals zero to infinity.

03:01 C -n plus two, n plus two, n plus 1, x to the n, minus n equals 1 to infinity.

03:13 Whoops.

03:17 C -n -n -x -to -the -n plus n equals zero to infinity for c -n -x -to -the -n.

03:28 All right, then the next thing before you can solve is they have to all start at the end.

03:32 The same in.

03:34 This one starts at zero.

03:35 This one starts at one.

03:36 This one starts at zero.

03:38 So we need the first and last one to start at one.

03:41 So first i'm going to plug in zero, see what we get.

03:45 And then the rest of it, i'll put back in as the series.

03:49 So you get c2 times two times one, times x to the zero, plus the rest of them, and equals one to infinity, cn plus two, n plus two, and plus one.

04:03 1 x to the n.

04:06 Now this one already starts at once.

04:08 I'm not going to do anything to it.

04:13 Okay, when you plug in 0 here you get 4c0 x to the 0.

04:18 Now plus all the rest of them.

04:26 All right.

04:26 So now all the series start with 1 and i have 2c2 plus 4 z 0 plus n equals 1 to infinity.

04:37 Cn plus 2 n plus 2 n plus 1 minus cn times n plus 4 cn x to the n equals 0 okay so this tells me that c2 equals minus 2 c0 and this all has to be 0 so cn plus 2 equals n minus 4 cn over n plus 2 n plus 1.

05:19 Okay, so i subtracted these over so it made it minus 4 and positive n, and they both have cn with them and then divided by n plus 2 n plus 1.

05:32 So to find solutions, first i'm going to let c0 equal 1 and c1 equals 0.

05:40 So let's talk about c1 equals 0.

05:42 We're going to use this formula to find the rest of them.

05:46 So c3 would be, i'm plugging in 1 for n.

05:53 So minus 3, c1 over 3 times 2.

05:59 But c1 is 0.

06:01 So in fact, what happens is all the odd ones turn out to be 0 here.

06:08 All right, so it's the even ones we're interested in...

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