Applying Series Solution to First-order ODEs
Although having been touched on in the last tutorial, here we are going to formally introduce the method of series solution that is to be applied to first-order ODEs. Given a general first-order linear ODE:
\begin{align}
\frac{dy}{dx} + P(x)y = Q(x) \tag{1}
\end{align}
we may propose the series solution of
\begin{align}
y(x) &= a_0 + a_1 x + a_2x^2 + a_3x^3 + \cdots = \sum_{n=0}^{\infty} a_nx^n \tag{2} \\
y'(x) &= a_1 + 2a_2x + 3a_3x^2 + \cdots = \sum_{n=1}^{\infty} n a_nx^{n-1} \tag{3}
\end{align}
in addition to its first-order derivative, as suggested previously. Substituting (2) and (3) into (1) and equating the coefficients will then produce a recurrence relation that can be used to compute the coefficients \( a_n \) successively. The complexity depends on the form of \( P(x) \): easy if it is a constant or just a single power of \(x\), difficult if it is other functions that have a Taylor series with multiple terms; and \( Q(x) \), which affects the algebraic process to solve \( a_n \) at each step. Finally, we will have to check for the analyticity of the series solution.
Example
Find the series solution to the first-order ODE
\begin{align}
\frac{dy}{dx} + y = \arctan x \tag{4}
\end{align}
Substituting (2) and (3) into (4) leads to
\begin{align}
\sum_{n=1}^{\infty} n a_nx^{n-1} + \sum_{n=0}^{\infty} a_nx^n &= x -\frac{x^3}{3} + \frac{x^5}{5} -\cdots \\
\sum_{n=0}^{\infty} (n+1) a_{n+1}x^n + \sum_{n=0}^{\infty} a_nx^n &= \sum_{m=0}^{\infty} \frac{(-1)^m x^{2m+1}}{2m+1} \tag{5}
\end{align}
where we have expressed \( \arctan x \) in terms of Taylor series as well, and replace \( n \) by \( n+1 \) in the first summation (and adjust the initial index as well). Unfortunately, due to the “on-and-off” pattern (only odd power terms present) in the Taylor series of \( Q(x) \), we don’t have a very clean recurrence relation or a simple closed-form solution. We will therefore only write down the first few terms of the series solution generated by (5):
\begin{align}
\begin{aligned}
a_1 + a_0 = 0 &\Rightarrow a_1 = -a_0 \\
2a_2 + a_1 = 1 &\Rightarrow a_2 = \frac{1 -a_1}{2} = \frac{1}{2} + \frac{a_0}{2} \\
3a_3 + a_2 = 0 &\Rightarrow a_3 = \frac{-a_2}{3} = -\frac{1}{6} -\frac{a_0}{6} \\
4a_4 + a_3 = -\frac{1}{3} &\Rightarrow a_4 = \frac{-1/3 -a_3}{4} = -\frac{1}{24} +\frac{a_0}{24} \\
\vdots
\end{aligned} \tag{6}
\end{align}
Hence
\begin{align}
y &= a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \cdots \\
&= a_0 -a_0x + (\frac{1}{2} + \frac{a_0}{2})x^2 \\
&\quad + (-\frac{1}{6} -\frac{a_0}{6})x^3 + (-\frac{1}{24} +\frac{a_0}{24})x^4 + \cdots \\
&= a_0(1 -x + \frac{1}{2}x^2 -\frac{1}{6}x^3 + \frac{1}{24}x^4 + \cdots) \\
&\quad+ (\frac{1}{2}x^2 -\frac{1}{6}x^3 -\frac{1}{24}x^4 +\cdots) \\
&= a_0e^{-x} + (\frac{1}{2}x^2 -\frac{1}{6}x^3 -\frac{1}{24}x^4 + \cdots) \tag{7}
\end{align}
With this form of series solution, an initial condition at \( x = 0 \) can be easily incorporated. For example, if \( y(0) = 3 \), then we just need to take \( a_0 = 3 \) as the other infinite series vanish.
Exercise
Solve the ODE
\begin{align}
\frac{dy}{dx} + xy = \frac{1}{1-x} \tag{8}
\end{align}
via the method of series solution.
Answer
Substituting (2) and (3) into (8), in addition to expanding \( 1/(1-x) \) into Taylor series, gives
\begin{align}
\sum_{n=1}^{\infty} n a_nx^{n-1} + x\sum_{n=0}^{\infty} a_nx^n &= 1+x+x^2+\cdots \\
\sum_{n=0}^{\infty} (n+1) a_{n+1}x^{n} + \sum_{n=0}^{\infty} a_nx^{n+1} &= \sum_{n=0}^{\infty} x^n \\
\sum_{n=0}^{\infty} (n+1) a_{n+1}x^{n} + \sum_{n=1}^{\infty} a_{n-1}x^n &= \sum_{n=0}^{\infty} x^n
\end{align}
where we have replaced \( n \) by \( n+1 \) and \( n-1 \) respectively in the two summations on L.H.S. Then by comparing the coefficients, the recurrence relation reads
\begin{align}
(n+1)a_{n+1} + a_{n-1} &= 1 \\
a_{n+1} &= \frac{1 -a_{n-1}}{n+1}
\end{align}
For \( n = 0 \), the second summation does not appear (for convenience we usually assume \(a_{-1} = 0\)), so it simply reduces to \( a_1 = 1 \). We subsequently generate some of the remaining terms:
\begin{align}
a_2 &= \frac{1 -a_0}{2} = \frac{1}{2} -\frac{a_0}{2} \\
a_3 &= \frac{1 -a_1}{3} = 0 \\
a_4 &= \frac{1 -a_2}{4} = \frac{1}{4}(\frac{1}{2} + \frac{a_0}{2}) = \frac{1}{8} + \frac{a_0}{8} \\
a_5 &= \frac{1 -a_3}{5} = \frac{1}{5} \\
a_6 &= \frac{1 -a_4}{6} = \frac{1}{6}(\frac{7}{8} -\frac{a_0}{8}) = \frac{7}{48} -\frac{a_0}{48} \\
\vdots
\end{align}
and hence the series solution is
\begin{align}
y &= a_0 + a_1x + a_2x^2 + a_3x^3 \\
&\quad + a_4x^4 + a_5x^5 + a_6x^6 + \cdots \\
&= a_0 + x + (\frac{1}{2} -\frac{a_0}{2})x^2 + 0 \\
&\quad + (\frac{1}{8} + \frac{a_0}{8})x^4 + \frac{1}{5}x^5 + (\frac{7}{48} -\frac{a_0}{48})x^6 + \cdots \\
&= a_0(1 -\frac{1}{2}x^2 + \frac{1}{8}x^4 -\frac{1}{48}x^6 + \cdots) \\
&\quad + (x + \frac{1}{2}x^2 + \frac{1}{8}x^4 + \frac{1}{5}x^5 + \frac{7}{48}x^6 + \cdots) \\
&= a_0e^{-x^2/2} + (x + \frac{1}{2}x^2 + \frac{1}{8}x^4 + \frac{1}{5}x^5 + \frac{7}{48}x^6 + \cdots)
\end{align}
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