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Bobbie Comstock

Answered question

2022-01-19

Solving a differential equation by power series.
I want to find the power series solutions about the origin of two linearly independent solutions of

w - z w = 0

.
Also, how do I show that these solutions are analytic?

Answer & Explanation

braodagxj

Beginner2022-01-19Added 38 answers

w (x) = a_{0} + a_{1} x + \frac{a_{2} x^{2}}{2!} + \frac{a_{3} x^{3}}{3!} + \dots

w (x) = a_{2} + a_{3} x + \frac{a_{4} x^{2}}{2!} + \frac{a_{5} x^{3}}{3!} + \dots

w - x w = 0

(a_{2} + a_{3} x + \frac{a_{4} x^{2}}{2!} + \frac{a_{5} x^{3}}{3!} + \dots)

- x (a_{0} + a_{1} x + \frac{a_{2} x^{2}}{2!} + \frac{a_{3} x^{3}}{3!} + \dots) = 0

a_{2} + (a_{3} - a_{0}) x + (\frac{a_{4}}{2!} - a_{1}) x^{2} + (\frac{a_{5}}{3!} - \frac{a_{2}}{2!}) x^{3} + \dots = 0

a_{2} = 0

a_{3} - a_{0} = 0

\frac{a_{4}}{2!} - a_{1} = 0

\frac{a_{5}}{3!} - \frac{a_{2}}{2!} = 0

[n > 2] \frac{a_{n}}{(n - 2)!} - \frac{a_{n - 3}}{(n - 3)!} = 0

a_{n} = (n - 2) a_{n - 3}

a_{0} = c_{1} a_{1} = c_{2} a_{2} = 0

a_{3} = a_{0} = c_{1}

a_{4} = 2 a_{1} = 2 c_{2}

a_{5} = 3 a_{2} = 0

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Bertha Jordan

Beginner2022-01-20Added 37 answers

Let's say that w(z) is one such solution and suppose that it has a power series, convergent in some disk around the origin, given by

w (z) = \sum_{n} a_{n} z^{n}

You just substitute that into the equation, and get a recurrence relation for the coefficients

a_{0}, a_{1}, a_{2}

,…. But, in fact, you will get a linear recurrence relation that relates coefficients

a_{2}, a_{3}

,… to

a_{0} and a_{1}

, so what you get is

w (z) = a_{0} \sum_{n} a_{n}^{(1)} z^{n} + a_{1} \sum_{n} a_{n}^{(2)} z^{n}

,
where

a^{(1)} and a^{(2)}

are two distinct linearly independent solutions to the recurrence relation (for some equations it is a little trickier than this). The two power series define the two linearly independent solutions of the ODE.

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