00:01
In this problem, we want to use the power series method to solve the following differential equation.
00:06
We have 1 minus x times y prime prime plus y equal to 0.
00:13
And we want to solve using a power series centered about x equal to 0.
00:20
So first, let's write our power series for y.
00:24
Let's put y equal to the sum of a n times x power of n for n ranging between 0 and infinity.
00:31
From here, we can differentiate twice and obtain y ''.
00:37
The sum of an times n times n -1 times x to the power of n -2 for n ranging between 2 to infinity.
00:57
Let's rewrite this sum starting at 0.
00:59
And we obtain the sum of a n plus 2 times n plus 2 times n plus 1 times x to the power of n.
01:14
Now let's write these results into a differential equation, and we find that the sum of a n plus 2 times n plus 2 times n plus 1 times x to the power of n, n ranging between 0 and infinity minus the sum of a n plus 2 times n plus 2 times n plus 1 times x power n plus 1 to this this term here that i distributed on the y prime plus the sum of a n times x power n all this must be equal to 0 so the goal is to try to group up these three terms into one single summation but the problem is this middle term that has an x to the power of n plus 1.
02:31
So let's rewrite this as the sum of a n plus 1 times n plus 1 times n times x to the power of n, for n ranging between 1 to infinity.
02:48
So now we have all of our series, which is a function of x to the power of n, but this middle one is still fully problematic because it starts at 1.
02:58
But what we can do here is to remove the first term from our series.
03:06
So when n is equal to, or excuse me, rather, let's rewrite the first and last series starting at 1 by removing the first term.
03:27
So when n is equal to 0 in this series, we obtain a2.
03:33
And now let's add the sum of a n plus 2 times n plus 2 times n plus 1 times x power of n for n ranging between 1 to infinity.
03:52
Similarly, in this last series, let's remove our first term, a naught, and then write the remaining of our series starting at n equal to 1.
04:09
So now we have a naught plus a2 plus the sum of a n plus 2 times n plus 2 times n plus 1 minus a n plus 1 times n plus 1 times n plus a n times x times n is equal to 0 for n ranging between 1 to infinity.
04:38
So to so i have equality here for all n.
04:43
A1 plus a2 must be equal to 0.
04:50
And this term here in brackets must also be equal to 0 for all n.
05:09
So now let's develop the first eight terms.
05:13
So when n is equal to 0, we obtain that a2 times 2.
05:33
Oh, excuse me...