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Zondrina

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## Main Question or Discussion Point

I did this quite a while ago, and I wanted to know something specific.

Suppose we have a well behaved, second order, linear, homogeneous, ordinary differential equation. Suppose further all points are ordinary, and so we can seek an ordinary solution about ##x_0 = 0##. Then the solution takes the form:

$$y = \sum_{n = 0}^{\infty} a_n (x - x_0)^n = \sum_{n = 0}^{\infty} a_n x^n = \sum_{k = 0}^{\infty} a_{2k} x^{2k} + \sum_{k = 0}^{\infty} a_{2k + 1} x^{2k + 1}$$

Where we may deduce ##a_n## from a recurrence relation on ##n##.

Now my question is, if I chose ##x_0 = (a \neq 0) \in \mathbb{R}## instead of ##x_0 = 0##, would the solution take the form:

$$y = \sum_{n = 0}^{\infty} a_n (x - x_0)^n = \sum_{n = 0}^{\infty} a_n (x - a)^n = \sum_{k = 0}^{\infty} a_{2k} (x - a)^{2k} + \sum_{k = 0}^{\infty} a_{2k + 1} (x - a)^{2k + 1}$$

?

Thank you.

Suppose we have a well behaved, second order, linear, homogeneous, ordinary differential equation. Suppose further all points are ordinary, and so we can seek an ordinary solution about ##x_0 = 0##. Then the solution takes the form:

$$y = \sum_{n = 0}^{\infty} a_n (x - x_0)^n = \sum_{n = 0}^{\infty} a_n x^n = \sum_{k = 0}^{\infty} a_{2k} x^{2k} + \sum_{k = 0}^{\infty} a_{2k + 1} x^{2k + 1}$$

Where we may deduce ##a_n## from a recurrence relation on ##n##.

Now my question is, if I chose ##x_0 = (a \neq 0) \in \mathbb{R}## instead of ##x_0 = 0##, would the solution take the form:

$$y = \sum_{n = 0}^{\infty} a_n (x - x_0)^n = \sum_{n = 0}^{\infty} a_n (x - a)^n = \sum_{k = 0}^{\infty} a_{2k} (x - a)^{2k} + \sum_{k = 0}^{\infty} a_{2k + 1} (x - a)^{2k + 1}$$

?

Thank you.