[Notes on Diffy Qs home] [PDF version] [Buy paperback on Amazon]

[next] [prev] [prev-tail] [tail] [up]

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 , , and are polynomials. We will try a solution of the form

and solve for the to try to obtain a solution deﬁned in some interval around .

The point is called an ordinary point if . That is, the functions

are deﬁned for near . If , then we say 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 , 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 term is a constant and hence disappears. We therefore get

We diﬀerentiate yet again to obtain (now the term disappears)

We reindex the series (replace with ) to obtain

Now we plug and into the diﬀerential equation

As 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 or was. They can be arbitrary. But once we pick and , then all other coeﬃcients are determined by the recurrence relation.

Let us see what the coeﬃcients must be. First, and are arbitrary

So we note that for even , that is we get

and for odd , that is 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. Suppose we wish to solve Airy’s equation^{2}, that is

near the point . Note that is an ordinary point.

We try

We diﬀerentiate twice (as above) to obtain

We plug into the equation

We reindex to make things easier to sum

Again is supposed to be 0 so ﬁrst we notice that and also

Now we jump in steps of three. First we notice that since we must have that, , , , etc…. In general .

The constants and are arbitrary and we obtain

For where is a multiple of , that is we notice that

For where , we notice

In other words, if we write down the series for we notice that it has two parts

We deﬁne

and write the general solution to the equation as . Notice from the power series that and . Also, and . Therefore is a solution that satisﬁes the initial conditions and .The functions and cannot be written in terms of the elementary functions that you know. See Figure 7.3 for the plot of the solutions and . These functions have many interesting properties. For example, they are oscillatory for negative (like solutions to ) and for positive they grow without bound (like solutions to ).

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 ^{3} is the equation

Let us ﬁnd a solution around the point . We try

We diﬀerentiate (as above) to obtain

Now we plug into the equation

As we have

This recurrence relation actually includes (which comes about from ). Again and are arbitrary.

Let us separate the even and odd coeﬃcients. We ﬁnd thatLet us write down the two series, one with the even powers and one with the odd.

We then writeWe also notice that if is a positive even integer, then is a polynomial as all the coeﬃcients in the series beyond a certain degree are zero. If is a positive odd integer, then is a polynomial. For example, if , then

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 coeﬃcient.

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 at the point .

Exercise 7.2.6 (Chebyshev’s equation of order ): a) Solve using power series methods at . b) For what is there a polynomial solution?

Exercise 7.2.8: a) Use power series methods to solve at the point . b) Use the solution to part a) to ﬁnd a solution for around the point .

Exercise 7.2.102 (challenging): We can also use power series methods in nonhomogeneous equations. a) Use power series methods to solve at the point . Hint: Recall the geometric series. b) Now solve for the initial condition , .

Exercise 7.2.103: Attempt to solve at using the power series method of this section ( is a singular point). Can you ﬁnd at least one solution? Can you ﬁnd more than one solution?

^{2}Named after the English mathematician Sir George Biddell Airy (1801–1892).

^{3}Named after the French mathematician Charles Hermite (1822–1901).