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Section 6.2 Ordinary Point Solution Example

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Subsection 6.2.1 Solve y″+2xy′+y=0 with a series around x=0

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The first thing we do is set

y(x)=∑n=0∞anxn.

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We can differentiate this like we would for any polynomial:

y′(x)=∑n=0∞nanxn−1,y″(x)=∑n=0∞n(n−1)anxn−2.

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

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And now, we substitute these into our ODE y″+2xy′+y=0:

0=∑n=0∞n(n−1)xn−2+2x∑n=0∞nanxn−1+∑n=0∞anxn=∑n=0∞n(n−1)xn−2+∑n=0∞2nanxn+∑n=0∞anxn=∑n=0∞n(n−1)anxn−2+∑n=0∞(2n+1)anxn

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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

∑n=0∞n(n−1)anxn−2

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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.

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This would get us

0=∑n=0∞n(n−1)anxn−2+∑n=0∞(2n+1)anxn=∑n=0∞(n+2)(n+1)an+2xn+∑n=0∞(2n+1)anxn=∑n=0∞[(n+2)(n+1)an+2+(2n+1)an]xn.

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In order for equality to hold, the coefficient of xn must equal 0, so our recurrence relation for an is

(n+2)(n+1)an+2+(2n+1)an=0for all n≥0.

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This is usually enough to solve an exercise, since it can be difficult to find an explicit expression (no an+2 or anything) for an.

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

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If we solve for an+2 in (n+2)(n+1)an+2+(2n+1)an=0, we get

an+2=−2n+1(n+2)(n+1)an.

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So, an+2 is proportional to an. Similarly, an is proportional to an−2 and so on. At the very end, note that all even terms a2k are proportional to a0 and all odd terms are proportional to a1.

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Since each a2k is proportional to a0, let's write a2k=α2ka0, and since each a2k+1 is proportional to a1, let's write a2k+1=β2k+1a0. Then, our solution y(x) becomes

y(x)=∑n=0∞anxn,=∑k=0∞a2kx2k+∑k=0∞a2k+1x2k+1,=a0∑k=0∞α2kx2k+a1∑k=0∞β2k+1x2k+1.

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This is actually very similar to the more familiar general solution y(x)=c1y1+c2y2, except with c1=a0, c2=a1, and

y1=∑k=0∞α2kx2k,y2=∑k=0∞β2k+1x2k+1.

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These choices of y1 and y2 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!

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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 \(a_n\text{.}\)