- #1

Saladsamurai

- 3,020

- 7

## Homework Statement

I am following along in an example problem and I am getting hung up on a step. We are seeking a power series solution of the DE:

[tex](x - 1)y'' + y' +2(x - 1)y = 0 \qquad(1)[/tex]

With the initial values [itex]y(4) = 5 \text{ and }y'(4) = 0[/itex]. We seek the solution in the form

[tex]y(x) = \sum_{n=0}^{\infty}a_n(x - x_o)^n\qquad(2)[/tex]Here it is convenient to let x

_{o}= 4. So we seek[tex]y(x) = \sum_{n=0}^{\infty}a_n(x - 4)^n\qquad(3)[/tex]

Here is the book text:

To put (1) in standard form we divide by x - 1 which yields

[tex]y'' + \frac{1}{x - 1}y' + 2y = 0 \qquad(4)[/tex]

Essentially we put (1) in the standard form of (4) so that we can test for analyticity and we find that p(x) and q(x) in [itex]y'' + p(x)y' +q(x)y = 0[/itex] are analytic at x

_{o}= 4.

Ok that's all great.

**Here is where I lose them**:

I am not sure what is happening here. Why are we expanding the coefficients of the DE (1) ? When we had a simple constant coefficient DE, the procedure was simple:To proceed we can either use the form (1) or (4). Since we are expanding each term in the differential equation about x = 4, we need to expand (x - 1) and 2(x - 1) if we use (1), or the 1/(x - 1) factor if we use (4). The former is easier since [itex] x - 1 = 3 + (x - 4)\qquad(5)[/itex] is merely a two-term Taylor series, whereas 1/(x - 1) is an infinite series ...

*i. assume y(x) takes the form of (2); ii. differentiate (2) as many times as necessary and plug back into DE; iii. tidy up.*

Now we have a DE with variable coefficients the procedure is changing a bit, but I am not seeing how.

Let's talk generally for a moment. If I have a DE of the form [itex]A_1(x)y''(x) + A_2(x)y'(x) + A_3(x)y = 0 \qquad(6)[/itex], and I seek the solution in the form of [itex]y(x) = \sum_{n=0}^{\infty}a_n(x - x_o)^n\qquad(2)[/itex], how does the procedure that I outlined above for the constant coefficient case (italicized) change for the case of (6) with the

*A*coefficients? Do I need to expand my

_{i}(x)*A*' s about x

_{i}(x)_{o}as well? If yes, why? (I have a feeling I know the answer, but would like verification. I feel like it only makes sense that my coefficients that depend on

*x*should 'track' my assumed

*y(x)*. I know that is not very rigorous, but it's all I am coming up with.)

Thanks for reading