5.6 The Power Series Method
A power series is an infinite series of the form \[(1)\hspace {2cm} \sum ^{\infty }_{n = 0} a_n \big (x - x_0\big )^n = a_0 + a_1\big (x -x_0\big ) + a_2\big (x -x_0\big )^2 + \cdots \cdots \cdots \cdots \]
If the series (1) converges for all \(x\) such that \(\big |x - x_0\big | <R\) and diverges for all \(x \ni \big |x -x_0\big | > R\), then \(R\) is called the radius of convergence of (1). It can be obtained as follows \[(a)\hspace {0.4cm} R = \frac {1}{\lim \limits _{n \rightarrow \infty } \sqrt [n]{\big |a_n\big |}}\hspace {0.5cm} \text {or}\hspace {0.5cm} (b)\hspace {0.4cm} R = \frac {1}{\lim \limits _{n \rightarrow \infty } \Big |\dfrac {a_{n+1}}{a_n}\Big |}\] provided these limits exists and are not zero. The interval over which the series converges is called the convergence interval.
If the convergence interval is infinite, then (1) converges for all \(x\).
Now suppose we have a linear ODE \[p(x,y) y'' + q(x)y' + r(x)y = 0\] Suppose \(p,q\) and \(r\) are polynomials , we will try the solution \[y(x) = \sum ^{\infty }_{k=0} a_k\big (x -x_0\big )^k\] and solve for \(a_k\) to obtain a solution defined in some interval around \(x_0\). The point \(x_0\) is called an ordinary (or regular) point if \(p(x_0) \neq 0\). That is \(\dfrac {q(x)}{p(x)}\) and \(\dfrac {r(x)}{p(x)}\) are defined for all \(x\) near \(x_0\). If \(p(x_0) = 0\) , then \(x_0\) is a singular point.
We first consider cases in which \(x_0\) is a regular point
Solution.
Let \(\displaystyle {y(x) = \sum ^{\infty } _{k = 0} a_k x^k}\)
\begin {align*} y'(x) & = \sum ^{\infty } _{k = 1}k a_k x^{k-1}\\\\ y''(x) & = \sum ^{\infty } _{k = 2}k(k-1)a_k x^{k-2} \end {align*}
Substituting in DE \[\sum ^{\infty } _{k = 2}k(k-1)a_k x^{k-2} -\sum ^{\infty } _{k = 0} a_k x^k = 0\]
We reindex the first series by simply replacing \(k + k + 2\). So, we get \[\sum ^{\infty }_{k + 2 = 2} (k + 2) \big [(k+2) -1\big ] a_{k + 2} \hspace {0.1cm} x^k - \sum ^{\infty }_{k = 0} a_k \hspace {0.1cm}x^k = 0\]
\[\sum ^{\infty }_{k = 0} (k+ 2 ) (k + 1) \hspace {0.1cm} a_{k + 2} \hspace {0.1cm} x^k - \sum ^{\infty }_{k = 0} a_k\hspace {0.1cm} x^k = 0\]
\[\sum ^{\infty }_{k=0}\Big [(k+2)(k+1)a_{k+2} - a_k\Big ] x^k = 0\]
Series \(= 0 \implies \) coefficients are zero
\[\implies (k+2)(k+1)a_{k+2} - a_k = 0\]
\[a_{k+2} = \frac {a_k}{(k+2)(k+1)}\hspace {0.3cm},\hspace {0.3cm} k = 0,1,2,\cdot \cdots \cdots \]
\begin {align*} k & = 0 \implies a_2 = \frac {a_0}{(2)(1)} = \frac {a_0}{2!}\\\\ k & = 1 \implies a_3 = \frac {a_1}{2(3)} = \frac {a_1}{3!}\\\\ k & = 2 \implies a_4 = \frac {a_2}{3(4)} = \frac {a_0}{1(2)(3)(4)} = \frac {a_0}{4!}\\\\ k & = 3 \implies a_5 = \frac {a_3}{4(5)} = \frac {a_1}{2(3)(4)(5)}=\frac {a_1}{5!}\\\\ k & = 4 \implies a_6 = \frac {a_4}{5(6)}= \frac {a_0}{1(2)(3)(4)(5)(6)}= \frac {a_0}{6!} \end {align*}
\[\implies \hspace {1cm} y(x) = \sum ^{\infty }_{k = 0} a_k\hspace {0.1cm}x^k = a_0 + a_1x + a_2 x^2 + a_3x^3+ \cdots \cdots \cdots \cdots \]
\[\text {So},\hspace {0.3cm} y(x) = a_0 + a_1x + \frac {a_0}{2!}x^2 + \frac {a_1}{3!}x^3 + \frac {a_0}{4!}x^4 + \frac {a_1}{5!}x^5 +\cdots \cdots \cdots \]
\[\therefore \hspace {0.5cm} y(x) = a_0\Bigg ( 1 + \frac {x^2}{2!}+ \frac {x^4}{4!} +\cdots \cdots \Bigg ) + a_1 \Bigg (x + \frac {x^3}{3!} + \frac {x^5}{5!}+ \cdots \cdots \Bigg )\]
\[y(x) = a_0 \sum ^{\infty }_{n = 0} \frac {x^{2n}}{\big (2n\big )!} + a_1 \sum ^{\infty }_{n=0} \frac {x^{2n+1}}{\big (2n+1\big )!}\]
\[\therefore \hspace {1cm} y(x) = a_0 \cosh (x) + a_1 \sinh (x)\]
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