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Section 6.2 Ordinary Point Solution Example

Subsection 6.2.1 Solve \(y'' + 2xy' + y = 0\) with a series around \(x=0\)

The first thing we do is set

\begin{align*} y(x) \amp = \sum_{n=0}^\infty a_nx^n. \end{align*}

We can differentiate this like we would for any polynomial:

\begin{align*} y'(x) \amp = \sum_{n=0}^\infty na_n x^{n-1}, \qquad y''(x) = \sum_{n=0}^\infty n(n-1)a_nx^{n-2}. \end{align*}

(Technically, \(y'\) should start with \(n=1\) and \(y''\) should start with \(n=2\) but that shouldn't matter all too much.)

And now, we substitute these into our ODE \(y'' + 2xy' + y = 0\text{:}\)

\begin{align*} 0 \amp = \sum_{n=0}^\infty n(n-1)x^{n-2} + 2x\sum_{n=0}^\infty na_n x^{n-1} + \sum_{n=0}^\infty a_nx^n\\ \amp = \sum_{n=0}^\infty n(n-1)x^{n-2} + \sum_{n=0}^\infty 2na_n x^n + \sum_{n=0}^\infty a_nx^n\\ \amp = \sum_{n=0}^\infty n(n-1)a_nx^{n-2} + \sum_{n=0}^\infty (2n+1)a_n x^n \end{align*}

We want to combine all the sums so that there's only one summation. To combine the two sums above, we need to use an “index shift” on the sum

\begin{gather*} \sum_{n=0}^\infty n(n-1)a_nx^{n-2} \end{gather*}

so that the powers of \(x\) match in each sum.

We bascially replace the index variable (which in the case of our sums is \(n\)) with something else.

Since we want to match the powers of \(x\text{,}\) let's write \(n = k+2\) in the first sum above. That way, we'd have

\begin{align*} \sum_{n=0}^\infty n(n-1)a_nx^{n-2} \amp = \sum_{k+2=0}^\infty (k+2)(k+2-1)a_{k+2}x^{k+2-2} = \sum_{k=-2}^\infty (k+2) (k+1) a_{k+2} x^k. \end{align*}

The \(k = -2\) is a little scary, but notice that \((k+2) (k+1) a_{k+2} x^k = 0\) when \(k = -2\) and when \(k = -1\text{.}\) So, we can rewrite the sum as

\begin{align*} \sum_{k=-2}^\infty (k+2) (k+1) a_{k+2} x^k \amp = \sum_{k=0}^\infty (k+2) (k+1) a_{k+2} x^k = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n, \end{align*}

where in the last step, we've just replaced all the \(k\)'s with \(n\)'s.

This would get us

\begin{align*} 0 \amp = \sum_{n=0}^\infty n(n-1)a_nx^{n-2} + \sum_{n=0}^\infty (2n+1)a_n x^n\\ \amp = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n + \sum_{n=0}^\infty (2n+1)a_n x^n\\ \amp = \sum_{n=0}^\infty \Big[ (n+2)(n+1) a_{n+2} + (2n+1)a_n \Big]x^n. \end{align*}

In order for equality to hold, the coefficient of \(x^n\) must equal \(0\text{,}\) so our recurrence relation for \(a_n\) is

\begin{align*} (n+2)(n+1) a_{n+2} + (2n+1)a_n \amp = 0 \qquad \text{for all $n \geq 0$}. \end{align*}

This is usually enough to solve an exercise, since it can be difficult to find an explicit expression (no \(a_{n+2}\) or anything) for \(a_n\text{.}\)

Something very important to mention here though, is that only even terms depend on even terms, and only odd terms depend on odd terms.

If we solve for \(a_{n+2}\) in \((n+2)(n+1) a_{n+2} + (2n+1)a_n = 0\text{,}\) we get

\begin{align*} a_{n+2} \amp = - \frac{2n+1}{(n+2)(n+1)} a_n. \end{align*}

So, \(a_{n+2}\) is proportional to \(a_n\text{.}\) Similarly, \(a_n\) is proportional to \(a_{n-2}\) and so on. At the very end, note that all even terms \(a_{2k}\) are proportional to \(a_0\) and all odd terms are proportional to \(a_1\text{.}\)

Since each \(a_{2k}\) is proportional to \(a_0\text{,}\) let's write \(a_{2k} = \alpha_{2k} a_0\text{,}\) and since each \(a_{2k+1}\) is proportional to \(a_1\text{,}\) let's write \(a_{2k+1} = \beta_{2k+1} a_0\text{.}\) Then, our solution \(y(x)\) becomes

\begin{align*} y(x) \amp = \sum_{n=0}^\infty a_nx^n,\\ \amp = \sum_{k=0}^\infty a_{2k}x^{2k} + \sum_{k=0}^\infty a_{2k+1}x^{2k+1},\\ \amp = a_0\sum_{k=0}^\infty \alpha_{2k} x^{2k} + a_1\sum_{k=0}^\infty \beta_{2k+1} x^{2k+1}. \end{align*}

This is actually very similar to the more familiar general solution \(y(x) = c_1y_1 + c_2y_2\text{,}\) except with \(c_1 = a_0\text{,}\) \(c_2 = a_1\text{,}\) and

\begin{align*} y_1 \amp = \sum_{k=0}^\infty \alpha_{2k} x^{2k}, \qquad y_2 = \sum_{k=0}^\infty \beta_{2k+1} x^{2k+1}. \end{align*}

These choices of \(y_1\) and \(y_2\) form a fundamental pair of solutions (take it as a given for now). Thus, the solution \(y(x)\) was actually the general solution for the ODE!

With these kinds of questions, we're often asked about the first few terms:

Remember our recurrence relation:

\begin{gather*} (n+2)(n+1) a_{n+2} + (2n+1)a_n = 0 \quad \implies \quad a_{n+2} = - \frac{2n+1}{(n+2)(n+1)} a_n. \end{gather*}

Now, we just need to plug in certain values of \(n\text{.}\) (Don't forget that \(a_0\) and \(a_1\) are arbitrary.)

\begin{align*} a_2 \amp = - \frac{2(0) + 1}{(0 + 2)(0 + 1)} a_0 = -\frac12 a_0,\\ a_3 \amp = - \frac{2(1) + 1}{(1 + 2)(1 + 1)} a_1 = -\frac12 a_1,\\ a_4 \amp = - \frac{2(2) + 1}{(2 + 2)(2 + 1)} a_2 = -\frac5{12} \cdot -\frac12 a_0 = \frac5{24} a_0,\\ a_5 \amp = - \frac{2(3) + 1}{(3 + 2)(3 + 1)} a_3 = -\frac7{20} \cdot -\frac12 a_1 = \frac7{40} a_1. \end{align*}

So, the first six terms of \(y(x)\) are

\begin{gather*} a_0 + a_1x - \frac12 a_0 x^2 - \frac12 x^3 + \frac5{24} a_0x^4 + \frac7{40} a_1 x^5. \end{gather*}

This is usually something that you'd rarely have to do. We have to consider different cases based on whether \(n = 2k\) is even or if \(n = 2k+1\) is odd.

Our recurrence relation tells us that \(a_n = f(n) a_{n-2}\) where

\begin{align*} f(n) \amp = - \frac{2(n-2)+1}{((n-2)+2)((n-2)+1)} = -\frac{2n-3}{n(n-1)}. \end{align*}

When \(n = 2k\) is even, we find that

\begin{align*} a_{2k} \amp = f(2k) a_{2k-2} = f(2k) f(2k-2) a_{2k-4}, \phantom{\frac12} \\ \amp = f(2k) f(2k-2) f(2k-4) \cdots f(4) f(2) a_0,\\ \amp = a_0 \prod_{s=1}^k f(2s), \\ \amp = a_0 \prod_{s=1}^k \left[-\frac{4s-3}{2s(2s-1)}\right], \\ \amp = a_0 \frac{(-1)^k}{(2k)!} \prod_{s=1}^k (4s-3). \end{align*}

When \(n = 2k+1\) is odd, we find that

\begin{align*} a_{2k+1} \amp = f(2k+1) a_{2k-1} = f(2k+1) f(2k-1) a_{2k-3}, \phantom{\frac12} \\ \amp = f(2k+1) f(2k-1) f(2k-3) \cdots f(5) f(3) a_1,\\ \amp = a_1 \prod_{s=1}^k f(2s+1),\\ \amp = a_1 \prod_{s=1}^k \left[-\frac{4s-1}{(2s+1)2s}\right],\\ \amp = a_1 \frac{(-1)^k}{(2k+1)!} \prod_{s=1}^k (4s-1). \end{align*}

Put together, we get the explicit form of \(a_n\text{.}\)