Power series solution method for differential equations - Revision history

Vipul: /* Examples */

2012-07-09T21:16:37Z

Examples

← Older revision		Revision as of 21:16, 9 July 2012
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	<math>a_0 \left(1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \dots \right) + a_1 \left(x + \frac{x^4}{4!} + \frac{x^7}{7!} + \dots \right) + a_2 \left(\frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \dots \right), \qquad a_0, a_1, a_2 \in \R</math>		<math>a_0 \left(1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \dots \right) + a_1 \left(x + \frac{x^4}{4!} + \frac{x^7}{7!} + \dots \right) + a_2 \left(\frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \dots \right), \qquad a_0, a_1, a_2 \in \R</math>

			Comparison with the other method for solving this differential equation: {{fillin}}

Vipul: /* Examples */

2012-07-09T21:15:57Z

Examples

← Older revision		Revision as of 21:15, 9 July 2012
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	We thus see that <math>a_0,a_1,a_2</math> can be varied freely, and once these are determined, all the coefficients are determined. Explicitly, we have <math>a_0 = a_3 = a_6 = \dots</math>, separately <math>a_1 = a_4 = a_7 = \dots</math>, and separately <math>a_2 = a_5 = a_8 = \dots</math>. Thus, the power series can be written as:		We thus see that <math>a_0,a_1,a_2</math> can be varied freely, and once these are determined, all the coefficients are determined. Explicitly, we have <math>a_0 = a_3 = a_6 = \dots</math>, separately <math>a_1 = a_4 = a_7 = \dots</math>, and separately <math>a_2 = a_5 = a_8 = \dots</math>. Thus, the power series can be written as:

	<math>a_0 \left(1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \dots \right) + a_1 \left(x + \frac{x^4}{4!} + \frac{x^7}{7!} + \dots \right) + a_2 \left(\frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \dots \right)</math>		<math>a_0 \left(1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \dots \right) + a_1 \left(x + \frac{x^4}{4!} + \frac{x^7}{7!} + \dots \right) + a_2 \left(\frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \dots \right), \qquad a_0, a_1, a_2 \in \R</math>

Vipul: Created page with "==Description of the method== This method aims to find power series for the solution functions to a differential equation. The general idea is as follows: # Assume t..."

2012-07-09T21:15:04Z

Created page with "==Description of the method== This method aims to find power series for the solution functions to a differential equation. The general idea is as follows: # Assume t..."

New page

==Description of the method==

This method aims to find [[power series]] for the solution functions to a [[differential equation]]. The general idea is as follows:

# Assume that the solution function has a [[power series]] that converges to it. It is typically helpful to write the power series as an exponential generating function, i.e., in the form <math>\sum_{k=0}^\infty a_k\frac{x^k}{k!}</math>.
# Compute the power series for the derivatives of the function, and then plug in the differential equation to obtain a recurrence relation between the <math>a_k</math>s.
# Solve this recurrence relation. In general, we expect that all the coefficients are determined by the first few, where the number of initial coefficients that we are free to specify equals the order of the differential equation. For an initial value problem, these first few coefficients can be determined.

==Examples==

Consider the differential equation where the independent variable is <math>x</math> and the dependent variable is <math>y</math>:

<math>y''' = y</math>

Suppose <math>y</math> has a power series:

<math>y = \sum_{k=0}^\infty a_k \frac{x^k}{k!}</math>

It is easy to see that:

<math>y''' = \sum_{k=0}^\infty a_{k+3} \frac{x^k}{k!}</math>

The equation <math>y''' = y</math> means that these power series are equal coefficient-wise, so we get:

<math>a_k = a_{k+3} \ \forall \ k \in \{ 0,1,2,\dots \}</math>

We thus see that <math>a_0,a_1,a_2</math> can be varied freely, and once these are determined, all the coefficients are determined. Explicitly, we have <math>a_0 = a_3 = a_6 = \dots</math>, separately <math>a_1 = a_4 = a_7 = \dots</math>, and separately <math>a_2 = a_5 = a_8 = \dots</math>. Thus, the power series can be written as:

<math>a_0 \left(1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \dots \right) + a_1 \left(x + \frac{x^4}{4!} + \frac{x^7}{7!} + \dots \right) + a_2 \left(\frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + \dots \right)</math>