Applications of Series: Learn It 3

Giselle Miller

Applications of Series: Learn It 3

Solving Differential Equations with Power Series

Consider the differential equation

[latex]{y}^{\prime }\left(x\right)=y[/latex].

This is a first-order separable equation with solution [latex]y=C{e}^{x}[/latex]. We can solve this using techniques from earlier chapters.

For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form [latex]y=\displaystyle\sum_{n=0}^{\infty}{c}_{n}{x}^{n}[/latex] and determine what the coefficients need to be.

In the next example, we consider an initial-value problem involving [latex]{y}^{\prime}=y[/latex] to illustrate the technique.

Use power series to solve the initial-value problem

[latex]{y}^{\prime }=y,y\left(0\right)=3[/latex].

Watch the following video to see the worked solution to the example above.

You can view the transcript for “6.4.4” here (opens in new window).

We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation:

[latex]{y}^{\prime }-xy=0[/latex]

is known as Airy’s equation. It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Use power series to solve

[latex]y^{\prime\prime}-xy=0[/latex]

with the initial conditions [latex]y\left(0\right)=a[/latex] and [latex]y^{\prime} \left(0\right)=b[/latex].

We look for a solution of the form

[latex]y=\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\cdots[/latex].

Differentiating this function term by term, we obtain

[latex]\begin{array}{ccc}\hfill {y}^{\prime }& =\hfill & {c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\cdots ,\hfill \\ \hfill y\text{{\'\'} }& =\hfill & 2\cdot 1{c}_{2}+3\cdot 2{c}_{3}x+4\cdot 3{c}_{4}{x}^{2}+\cdots .\hfill \end{array}[/latex]

If y satisfies the equation [latex]y^{\prime\prime}=xy[/latex], then

[latex]2\cdot 1{c}_{2}+3\cdot 2{c}_{3}x+4\cdot 3{c}_{4}{x}^{2}+\cdots =x\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\cdots \right)[/latex].

Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

[latex]\begin{array}{c}2\cdot 1{c}_{2}=0,\hfill \\ 3\cdot 2{c}_{3}={c}_{0},\hfill \\ 4\cdot 3{c}_{4}={c}_{1},\hfill \\ 5\cdot 4{c}_{5}={c}_{2},\hfill \\ \hfill \vdots.\hfill \end{array}[/latex]

More generally, for [latex]n\ge 3[/latex], we have [latex]n\cdot \left(n - 1\right){c}_{n}={c}_{n - 3}[/latex]. In fact, all coefficients can be written in terms of [latex]{c}_{0}[/latex] and [latex]{c}_{1}[/latex]. To see this, first note that [latex]{c}_{2}=0[/latex]. Then

[latex]\begin{array}{}\\ \\ {c}_{3}=\frac{{c}_{0}}{3\cdot 2},\hfill \\ {c}_{4}=\frac{{c}_{1}}{4\cdot 3}.\hfill \end{array}[/latex]

For [latex]{c}_{5},{c}_{6},{c}_{7}[/latex], we see that

[latex]\begin{array}{}\\ \\ {c}_{5}=\frac{{c}_{2}}{5\cdot 4}=0,\hfill \\ {c}_{6}=\frac{{c}_{3}}{6\cdot 5}=\frac{{c}_{0}}{6\cdot 5\cdot 3\cdot 2},\hfill \\ {c}_{7}=\frac{{c}_{4}}{7\cdot 6}=\frac{{c}_{1}}{7\cdot 6\cdot 4\cdot 3}.\hfill \end{array}[/latex]

Therefore, the series solution of the differential equation is given by

[latex]y={c}_{0}+{c}_{1}x+0\cdot {x}^{2}+\frac{{c}_{0}}{3\cdot 2}{x}^{3}+\frac{{c}_{1}}{4\cdot 3}{x}^{4}+0\cdot {x}^{5}+\frac{{c}_{0}}{6\cdot 5\cdot 3\cdot 2}{x}^{6}+\frac{{c}_{1}}{7\cdot 6\cdot 4\cdot 3}{x}^{7}+\cdots[/latex].

The initial condition [latex]y\left(0\right)=a[/latex] implies [latex]{c}_{0}=a[/latex]. Differentiating this series term by term and using the fact that [latex]{y}^{\prime }\left(0\right)=b[/latex], we conclude that [latex]{c}_{1}=b[/latex]. Therefore, the solution of this initial-value problem is

[latex]y=a\left(1+\frac{{x}^{3}}{3\cdot 2}+\frac{{x}^{6}}{6\cdot 5\cdot 3\cdot 2}+\cdots \right)+b\left(x+\frac{{x}^{4}}{4\cdot 3}+\frac{{x}^{7}}{7\cdot 6\cdot 4\cdot 3}+\cdots \right)[/latex].