## Section 7.2 Series solutions of linear second order ODEs

¶*Note: 1 or 1.5 lecture, §8.2 in [EP], §5.2 and §5.3 in [BD]*

Suppose we have a linear second order homogeneous ODE of the form

Suppose that \(p(x)\text{,}\) \(q(x)\text{,}\) and \(r(x)\) are polynomials. We will try a solution of the form

and solve for the \(a_k\) to try to obtain a solution defined in some interval around \(x_0\text{.}\)

The point \(x_0\) is called an *ordinary point* if \(p(x_0) \not= 0\text{.}\) That is, the functions

are defined for \(x\) near \(x_0\text{.}\) If \(p(x_0) = 0\text{,}\) then we say \(x_0\) is a *singular point*. Handling singular points is harder than ordinary points and so we now focus only on ordinary points.

###### Example 7.2.1.

Let us start with a very simple example

Let us try a power series solution near \(x_0 = 0\text{,}\) which is an ordinary point. Every point is an ordinary point in fact, as the equation is constant coefficient. We already know we should obtain exponentials or the hyperbolic sine and cosine, but let us pretend we do not know this.

We try

If we differentiate, the \(k=0\) term is a constant and hence disappears. We therefore get

We differentiate yet again to obtain (now the \(k=1\) term disappears)

We reindex the series (replace \(k\) with \(k+2\)) to obtain

Now we plug \(y\) and \(y''\) into the differential equation

As \(y'' - y\) is supposed to be equal to 0, we know that the coefficients of the resulting series must be equal to 0. Therefore,

The above equation is called a *recurrence relation* for the coefficients of the power series. It did not matter what \(a_0\) or \(a_1\) was. They can be arbitrary. But once we pick \(a_0\) and \(a_1\text{,}\) then all other coefficients are determined by the recurrence relation.

Let us see what the coefficients must be. First, \(a_0\) and \(a_1\) are arbitrary

So we note that for even \(k\text{,}\) that is \(k=2n\) we get

and for odd \(k\text{,}\) that is \(k=2n+1\) we have

Let us write down the series

We recognize the two series as the hyperbolic sine and cosine. Therefore,

Of course, in general we will not be able to recognize the series that appears, since usually there will not be any elementary function that matches it. In that case we will be content with the series.

###### Example 7.2.2.

Let us do a more complex example. Consider *Airy's equation*^{ 1 }:

near the point \(x_0 = 0\text{.}\) Note that \(x_0 = 0\) is an ordinary point.

We try

We differentiate twice (as above) to obtain

We plug \(y\) into the equation

We reindex to make things easier to sum

Again \(y''-xy\) is supposed to be 0, so \(a_2 = 0\text{,}\) and

We jump in steps of three. First, since \(a_2 = 0\) we must have , \(a_5 = 0\text{,}\) \(a_8 = 0\text{,}\) \(a_{11}=0\text{,}\) etc. In general, \(a_{3n+2} = 0\text{.}\)

The constants \(a_0\) and \(a_1\) are arbitrary and we obtain

For \(a_k\) where \(k\) is a multiple of \(3\text{,}\) that is \(k=3n\) we notice that

For \(a_k\) where \(k = 3n+1\text{,}\) we notice

In other words, if we write down the series for \(y\text{,}\) it has two parts

We define

and write the general solution to the equation as \(y(x)= a_0 y_1(x) + a_1 y_2(x)\text{.}\) If we plug in \(x=0\) into the power series for \(y_1\) and \(y_2\text{,}\) we find \(y_1(0) = 1\) and \(y_2(0) = 0\text{.}\) Similarly, \(y_1'(0) = 0\) and \(y_2'(0) = 1\text{.}\) Therefore \(y = a_0 y_1 + a_1 y_2\) is a solution that satisfies the initial conditions \(y(0) = a_0\) and \(y'(0) = a_1\text{.}\)

The functions \(y_1\) and \(y_2\) cannot be written in terms of the elementary functions that you know. See Figure 8.2.3 for the plot of the solutions \(y_1\) and \(y_2\text{.}\) These functions have many interesting properties. For example, they are oscillatory for negative \(x\) (like solutions to \(y''+y=0\)) and for positive \(x\) they grow without bound (like solutions to \(y''-y=0\)).

Sometimes a solution may turn out to be a polynomial.

###### Example 7.2.3.

Let us find a solution to the so-called *Hermite's equation of order \(n\)*^{ 2 }:

Let us find a solution around the point \(x_0 = 0\text{.}\) We try

We differentiate (as above) to obtain

Now we plug into the equation

As \(y''-2xy'+2ny = 0\) we have

This recurrence relation actually includes \(a_2 = -na_0\) (which comes about from \(2a_2+2na_0 = 0\)). Again \(a_0\) and \(a_1\) are arbitrary.

Let us separate the even and odd coefficients. We find that

Let us write down the two series, one with the even powers and one with the odd.

We then write

We remark that if \(n\) is a positive even integer, then \(y_1(x)\) is a polynomial as all the coefficients in the series beyond a certain degree are zero. If \(n\) is a positive odd integer, then \(y_2(x)\) is a polynomial. For example, if \(n=4\text{,}\) then

### Subsection 7.2.1 Exercises

In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the \(k^{\text{th}}\) coefficient.

###### Exercise 7.2.1.

Use power series methods to solve \(y''+y = 0\) at the point \(x_0 = 1\text{.}\)

###### Exercise 7.2.2.

Use power series methods to solve \(y''+4xy = 0\) at the point \(x_0 = 0\text{.}\)

###### Exercise 7.2.3.

Use power series methods to solve \(y''-xy = 0\) at the point \(x_0 = 1\text{.}\)

###### Exercise 7.2.4.

Use power series methods to solve \(y''+x^2y = 0\) at the point \(x_0 = 0\text{.}\)

###### Exercise 7.2.5.

The methods work for other orders than second order. Try the methods of this section to solve the first order system \(y'-xy = 0\) at the point \(x_0 = 0\text{.}\)

###### Exercise 7.2.6. Chebyshev's equation of order \(p\).

Solve \((1-x^2)y''-xy' + p^2y = 0\) using power series methods at \(x_0=0\text{.}\)

For what \(p\) is there a polynomial solution?

###### Exercise 7.2.7.

Find a polynomial solution to \((x^2+1) y''-2xy'+2y = 0\) using power series methods.

###### Exercise 7.2.8.

Use power series methods to solve \((1-x)y''+y = 0\) at the point \(x_0 = 0\text{.}\)

Use the solution to part a) to find a solution for \(xy''+y=0\) around the point \(x_0=1\text{.}\)

###### Exercise 7.2.101.

Use power series methods to solve \(y'' + 2 x^3 y = 0\) at the point \(x_0 = 0\text{.}\)

\(a_2 = 0\text{,}\) \(a_3 = 0\text{,}\) \(a_4 = 0\text{,}\) recurrence relation (for \(k \geq 5\)): \(a_k
= \frac{- 2 a_{k-5}}{k(k-1)}\text{,}\) so:

\(y(x) = a_0 + a_1 x -\frac{a_0}{10} x^5 - \frac{a_1}{15} x^6
+ \frac{a_0}{450} x^{10} + \frac{a_1}{825} x^{11}
- \frac{a_0}{47250} x^{15} - \frac{a_1}{99000} x^{16}
+
\cdots\)

###### Exercise 7.2.102.

*(challenging)* Power series methods also work for nonhomogeneous equations.

Use power series methods to solve \(y'' - x y = \frac{1}{1-x}\) at the point \(x_0 = 0\text{.}\) Hint: Recall the geometric series.

Now solve for the initial condition \(y(0)=0\text{,}\) \(y'(0) = 0\text{.}\)

a) \(a_2 = \frac{1}{2}\text{,}\) and for \(k \geq 1\) we have \(a_k = \frac{a_{k-3} + 1}{k(k-1)}\text{,}\) so

\(y(x) = a_0 + a_1 x + \frac{1}{2} x^2
+ \frac{a_0 + 1}{6} x^3
+ \frac{a_1 + 1}{12} x^4
+ \frac{3}{40} x^5
+ \frac{a_0 + 2}{30} x^6
+ \frac{a_1 + 2}{42} x^7
+ \frac{5}{112} x^8
+ \frac{a_0 + 3}{72} x^9
+ \frac{a_1 + 3}{90} x^{10} +
\cdots\)

b) \(y(x) = \frac{1}{2} x^2
+ \frac{1}{6} x^3
+ \frac{1}{12} x^4
+ \frac{3}{40} x^5
+ \frac{1}{15} x^6
+ \frac{1}{21} x^7
+ \frac{5}{112} x^8
+ \frac{1}{24} x^9
+ \frac{1}{30} x^{10} +
\cdots\)

###### Exercise 7.2.103.

Attempt to solve \(x^2 y'' - y = 0\) at \(x_0 = 0\) using the power series method of this section (\(x_0\) is a singular point). Can you find at least one solution? Can you find more than one solution?

Applying the method of this section directly we obtain \(a_k = 0\) for all \(k\) and so \(y(x) = 0\) is the only solution we find.