9.1: Ordinary Points
If \(x_{0}\) is an ordinary point of \(Equation \ref{9.1}\) , then it is possible to determine two power series (i.e., Taylor series) solutions for \(y=y(x)\) centered at \(x=x_{0}\) . We illustrate the method of solution by solving two examples, with \(x_{0}=0\) .
Example: Find the general solution of \(y^{\prime \prime}+y=0\) .
By now, you should know that the general solution is \(y(x)=a_{0} \cos x+a_{1} \sin x\) , with \(a_{0}\) and \(a_{1}\) constants. To find a power series solution about the point \(x_{0}=0\) , we write
\[y(x)=\sum_{n=0}^{\infty} a_{n} x^{n} \nonumber \]
and upon differentiating term-by-term
\[y^{\prime}(x)=\sum_{n=1}^{\infty} n a_{n} x^{n-1}, \nonumber \]
and
\[y^{\prime \prime}(x)=\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2} \nonumber \]
Substituting the power series for \(y\) and its derivatives into the differential equation to be solved, we obtain
\[\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}+\sum_{n=0}^{\infty} a_{n} x^{n}=0 \nonumber \]
The power-series solution method requires combining the two sums on the lefthand-side of Equation \ref{9.2} into a single power series in \(x\) . To shift the exponent of \(x^{n-2}\) in the first sum upward by two to obtain \(x^{n}\) , we need to shift the summation index downward by two; that is,
\[\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1) a_{n+2} x^{n} . \nonumber \]
We can then combine the two sums in Equation \ref{9.2} to obtain
\[\sum_{n=0}^{\infty}\left((n+2)(n+1) a_{n+2}+a_{n}\right) x^{n}=0 . \nonumber \]
For Equation \ref{9.3} to be satisfied, the coefficient of each power of \(x\) must vanish separately. (This can be proved by setting \(x=0\) after successive differentiation.) We therefore obtain the recurrence relation
\[a_{n+2}=-\frac{a_{n}}{(n+2)(n+1)}, \quad n=0,1,2, \ldots \nonumber \]
We observe that even and odd coefficients decouple. We thus obtain two independent sequences starting with first term \(a_{0}\) or \(a_{1}\) . Developing these sequences, we have for the sequence beginning with \(a_{0}\) :
\[\begin{aligned} &a_{0} \\ &a_{2}=-\frac{1}{2} a_{0} \\ &a_{4}=-\frac{1}{4 \cdot 3} a_{2}=\frac{1}{4 \cdot 3 \cdot 2} a_{0} \\ &a_{6}=-\frac{1}{6 \cdot 5} a_{4}=-\frac{1}{6 !} a_{0} \end{aligned} \nonumber \]
and the general coefficient in this sequence for \(n=0,1,2, \ldots\) is
\[a_{2 n}=\frac{(-1)^{n}}{(2 n) !} a_{0} . \nonumber \]
Also, for the sequence beginning with \(a_{1}\) :
\[\begin{aligned} &a_{1} \\ &a_{3}=-\frac{1}{3 \cdot 2} a_{1} \\ &a_{5}=-\frac{1}{5 \cdot 4} a_{3}=\frac{1}{5 \cdot 4 \cdot 3 \cdot 2} a_{1}, \\ &a_{7}=-\frac{1}{7 \cdot 6} a_{5}=-\frac{1}{7 !} a_{1} \end{aligned} \nonumber \]
and the general coefficient in this sequence for \(n=0,1,2, \ldots\) is
\[a_{2 n+1}=\frac{(-1)^{n}}{(2 n+1) !} a_{1} . \nonumber \]
Using the principle of superposition, the general solution is therefore
\[\begin{aligned} y(x) &=a_{0} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n) !} x^{2 n}+a_{1} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1} \\ &=a_{0}\left(1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\ldots\right)+a_{1}\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots\right) \\ &=a_{0} \cos x+a_{1} \sin x \end{aligned} \nonumber \]
as expected.
In our next example, we will solve the Airy’s Equation. This differential equation arises in the study of optics, fluid mechanics, and quantum mechanics.
Example: Find the general solution of \(y^{\prime \prime}-x y=0\) .
With
\[y(x)=\sum_{n=0}^{\infty} a_{n} x^{n} \nonumber \]
the differential equation becomes
\[\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}-\sum_{n=0}^{\infty} a_{n} x^{n+1}=0 \nonumber \]
We shift the first sum to \(x^{n+1}\) by shifting the exponent up by three, i.e.,
\[\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}=\sum_{n=-1}^{\infty}(n+3)(n+2) a_{n+3} x^{n+1} \nonumber \]
When combining the two sums in Equation \ref{9.4}, we separate out the extra \(n=-1\) term in the first sum given by \(2 a_{2}\) . Therefore, \(Equation \ref{9.4}\) becomes
\[2 a_{2}+\sum_{n=0}^{\infty}\left((n+3)(n+2) a_{n+3}-a_{n}\right) x^{n+1}=0 . \nonumber \]
Setting coefficients of powers of \(x\) to zero, we first find \(a_{2}=0\) , and then obtain the recursion relation
\[a_{n+3}=\frac{1}{(n+3)(n+2)} a_{n} \nonumber \]
Three sequences of coefficients-those starting with either \(a_{0}, a_{1}\) or \(a_{2}\) -decouple. In particular the three sequences are
\[\begin{aligned} &a_{0}, a_{3}, a_{6}, a_{9}, \ldots \\ &a_{1}, a_{4}, a_{7}, a_{10}, \ldots \\ &a_{2}, a_{5}, a_{8}, a_{11} \ldots \end{aligned} \nonumber \]
Since \(a_{2}=0\) , we find immediately for the last sequence
\[a_{2}=a_{5}=a_{8}=a_{11}=\cdots=0 . \nonumber \]
We compute the first four nonzero terms in the power series with coefficients corresponding to the first two sequences. Starting with \(a_{0}\) , we have
\[\begin{aligned} &a_{0} \\ &a_{3}=\frac{1}{3 \cdot 2} a_{0} \\ &a_{6}=\frac{1}{6 \cdot 5 \cdot 3 \cdot 2} a_{0} \\ &a_{9}=\frac{1}{9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} a_{0} \end{aligned} \nonumber \]
and starting with \(a_{1}\) ,
\[\begin{aligned} a_{1} \\ a_{4} &=\frac{1}{4 \cdot 3} a_{1} \\ a_{7} &=\frac{1}{7 \cdot 6 \cdot 4 \cdot 3} a_{1} \\ a_{10} &=\frac{1}{10 \cdot 9 \cdot 7 \cdot 6 \cdot 4 \cdot 3} a_{1} \end{aligned} \nonumber \]
The general solution for \(y=y(x)\) , can therefore be written as
\[\begin{aligned} y(x) &=a_{0}\left(1+\frac{x^{3}}{6}+\frac{x^{6}}{180}+\frac{x^{9}}{12960}+\ldots\right)+a_{1}\left(x+\frac{x^{4}}{12}+\frac{x^{7}}{504}+\frac{x^{10}}{45360}+\ldots\right) \\ &=a_{0} y_{0}(x)+a_{1} y_{1}(x) \end{aligned} \nonumber \]
Suppose we would like to graph the solutions \(y=y_{0}(x)\) and \(y=y_{1}(x)\) versus \(x\) by solving the differential equation \(y^{\prime \prime}-x y=0\) numerically. What initial conditions should we use? Clearly, \(y=y_{0}(x)\) solves the ode with initial values \(y(0)=1\) and \(y^{\prime}(0)=0\) , while \(y=y_{1}(x)\) solves the ode with initial values \(y(0)=0\) and \(y^{\prime}(0)=1\) .
The numerical solutions, obtained using MATLAB, are shown in Fig. 9.1. Note that the solutions oscillate for negative \(x\) and grow exponentially for positive \(x\) . This can be understood by recalling that \(y^{\prime \prime}+y=0\) has oscillatory sine and cosine solutions and \(y^{\prime \prime}-y=0\) has exponential hyperbolic sine and cosine solutions.