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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

 ′′ ′ p(x)y + q (x)y + r(x)y = 0.

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

 ∑∞ k y = ak(x − x0) k=0

and solve for the ak to try to obtain a solution defined in some interval around x0 .

The point x 0 is called an ordinary point if p (x ) ⇔ 0 0 . That is, the functions

q(x) and r(x) p(x) p(x)

are defined for x near x 0 . If p(x0) = 0 , then we say x0 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

y′′ − y = 0.

Let us try a power series solution near x0 = 0 , 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

 ∞ ∑ k y = akx . k=0

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

 ∑∞ y′ = kakxk−1. k=1

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

 ∞∑ y′′ = k(k − 1)akxk−2. k=2

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

 ∑∞ y′′ = (k + 2)(k + 1)a xk. k+2 k=0

Now we plug y and  ′′ y into the differential equation

 ( ∞ ) ( ∞ ) ′′ ∑ k ∑ k 0 = y − y = (k + 2)(k + 1)ak+2x − akx ∞k=0 k=0 ∑ ( k k) = (k + 2)(k + 1)ak+2x − akx k=0 ∞∑ ( ) k = (k + 2)(k + 1)ak+2 − ak x . k=0

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,

 ak (k + 2)(k + 1)ak+2 − ak = 0, or ak+2 =-------------. (k + 2)(k + 1)

The above equation is called a recurrence relation for the coefficients of the power series. It did not matter what a0 or a 1 was. They can be arbitrary. But once we pick a0 and a1 , then all other coefficients are determined by the recurrence relation.

Let us see what the coefficients must be. First, a0 and a1 are arbitrary

 a0- --a1-- --a2-- ---a0--- --a3-- ----a1----- a2 = 2 , a 3 = (3)(2), a4 = (4)(3) = (4 )(3)(2), a5 = (5)(4 ) = (5)(4)(3)(2), ...

So we note that for even k , that is k = 2n we get

a = a = -a0--, k 2n (2n)!

and for odd k , that is k = 2n + 1 we have

 ---a1---- ak = a2n+1 = (2n + 1)!.

Let us write down the series

 ∑∞ ∑∞ ( ) ∑∞ ∑∞ y = akxk = -a0-x2n + ---a1---x2n+1 = a0 --1--x2n + a1 ----1----x2n+1. k=0 n=0 (2n )! (2n + 1)! n=0(2n)! n=0(2n + 1)!

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

y = a co sh x + a sin hx. 0 1

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 equation2, that is

y′′ − xy = 0,

near the point x 0 = 0 . Note that x0 = 0 is an ordinary point.

We try

 ∞∑ y = akxk. k=0

We differentiate twice (as above) to obtain

 ∞∑ y′′ = k(k − 1)akxk−2. k=2

We plug y into the equation

 (∑∞ ) (∑∞ ) 0 = y′′ − xy = k(k − 1)akxk−2 − x akxk k=2 k=0 (∑∞ ) (∑∞ ) = k(k − 1)akxk−2 − akxk+1 . k=2 k=0

We reindex to make things easier to sum

 ( ∑∞ ) (∑∞ ) 0 = y′′ − xy = 2a + (k + 2)(k + 1)a xk − a xk . 2 k=1 k+2 k=1 k−1 ∞ ∑ ( ) k = 2a2 + (k + 2)(k + 1)ak+2 − ak−1 x . k=1

Again y′′ − xy is supposed to be 0 so first we notice that a2 = 0 and also

 -----ak−1----- (k + 2)(k + 1)ak+2 − ak−1 = 0, or ak+2 = (k + 2)(k + 1).

Now we jump in steps of three. First we notice that since a = 0 2 we must have that, a = 0 5 , a = 0 8 , a11 = 0 , etc…. In general a3n+2 = 0 .

The constants a0 and a 1 are arbitrary and we obtain

 a a a a a a a3 = ---0-, a4 = --1--, a 6 =---3-- = -----0-----, a7 = ---4--= -----1----, ... (3)(2 ) (4 )(3) (6)(5) (6)(5 )(3)(2) (7)(6 ) (7)(6)(4)(3)

For a k where k is a multiple of 3 , that is k = 3n we notice that

 a a3n =-------------0------------. (2)(3)(5)(6)⋅⋅⋅(3n − 1)(3n)

For ak where k = 3n + 1 , we notice

 ------------a1------------ a3n+1 = (3)(4)(6 )(7)⋅⋅⋅(3n)(3n + 1).

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

 ( ) y = a 0 + a0x3 +-a0-x6 + ⋅⋅⋅ +-----------a0------------x3n + ⋅⋅⋅ 6 180 (2)(3)(5 )(6)⋅⋅⋅(3n − 1)(3n ) ( a1 a 1 a1 ) + a1x +---x4 + ---x7 + ⋅⋅⋅ +-------------------------x3n+1 + ⋅⋅⋅ ( 1 2 504 (3)(4 )(6)(7)⋅⋅⋅(3n)(3n + 1) ) 1-3 -1--6 ------------1-------------3n = a0 1 + 6x + 180x + ⋅⋅⋅ + (2)(3 )(5)(6)⋅⋅⋅(3n − 1 )(3n)x + ⋅⋅⋅ ( ) + a x + 1-x4 + -1--x7 + ⋅⋅⋅ +-----------1-------------x3n+1 + ⋅⋅⋅ . 1 12 504 (3)(4)(6 )(7)⋅⋅⋅(3n)(3n + 1)

We define

y (x) = 1 + 1x3 +-1--x6 + ⋅⋅⋅ +-----------1-------------x3n + ⋅⋅⋅, 1 6 180 (2)(3)(5 )(6)⋅⋅⋅(3n − 1)(3n ) 1 1 1 y2(x) = x +--x 4 +----x7 + ⋅⋅⋅ +-------------------------x3n+1 + ⋅⋅⋅, 12 504 (3)(4)(6)(7) ⋅⋅⋅(3n)(3n + 1)
and write the general solution to the equation as y(x ) = a0y1(x) + a1y2(x) . Notice from the power series that y1(0) = 1 and y2(0) = 0 . Also, y′(0) = 0 1 and y′(0) = 1 2 . Therefore y (x) is a solution that satisfies the initial conditions y(0) = a0 and  ′ y (0) = a1 .


PIC

Figure 7.3: The two solutions y1 and y2 to Airy’s equation.


The functions y1 and y2 cannot be written in terms of the elementary functions that you know. See Figure 7.3 for the plot of the solutions y1 and y2 . 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 3 is the equation

y′′ − 2xy′ + 2ny = 0.

Let us find a solution around the point x0 = 0 . We try

 ∞∑ k y = akx . k=0

We differentiate (as above) to obtain

 ∞∑ y′ = kakxk−1, k=1 ∞∑ y′′ = k(k − 1)a xk−2. k k=2

Now we plug into the equation

 ( ∞∑ ) (∑∞ ) (∑∞ ) 0 = y′′ − 2xy ′ + 2ny = k(k − 1)akxk−2 − 2x kakxk−1 + 2n akxk k=2 k=1 k=0 ( ∞∑ ) (∑∞ ) (∑∞ ) = k(k − 1)akxk−2 − 2kakxk + 2nakxk k=2 k=1 k=0 ( ∑∞ ) (∑ ∞ ) ( ∑∞ ) = 2a + (k + 2)(k + 1)a xk − 2ka xk + 2na + 2na xk 2 k=1 k+2 k=1 k 0 k=1 k ∞ ∑ ( ) k = 2a2 + 2na 0 + (k + 2)(k + 1 )ak+2 − 2kak + 2nak x . k=1

As y′′ − 2xy′ + 2ny = 0 we have

 (2k − 2n) (k + 2)(k + 1)ak+2 + (−2k + 2n)ak = 0, or ak+2 =-------------ak. (k + 2)(k + 1)

This recurrence relation actually includes a2 = − na0 (which comes about from 2a 2 + 2na 0 = 0 ). Again a0 and a1 are arbitrary.

 − 2n 2(1 − n) a2 =------a0, a3 = -------a1, (2)(1) (3)(2) 2-(2-−-n) 22(2 −-n-)(−-n) a4 = (4)(3) a2 = (4)(3 )(2)(1) a0, 2 a = 2-(3-−-n)a = 2-(3 −-n-)(1-−-n)a , ... 5 (5)(4) 3 (5)(4 )(3)(2) 1
Let us separate the even and odd coefficients. We find that
 2m-(−-n)(2-−-n)⋅⋅⋅(2m-−-2-−-n) a2m = (2m )! , m a 2m+1 = 2--(1-−-n)(3 −-n)⋅⋅⋅(2m-−-1-−-n). (2m + 1)!

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

 2(−n) 22(−n)(2 − n ) 23(−n )(2 − n)(4 − n) y1(x) = 1 + -----x2 + -------------x4 +-------------------x6 + ⋅⋅⋅, 2 ! 4! 6! 2(1 −-n) 3 22(1 −-n-)(3-−-n) 5 23(1-−-n)(3 −-n)(5-− n-) 7 y2(x) = x + 3 ! x + 5 ! x + 7! x + ⋅⋅⋅.
We then write
y(x) = a0y1(x) + a1y2(x).

We also notice that if n is a positive even integer, then y1(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 y2(x) is a polynomial. For example, if n = 4 , then

 2 y1(x) = 1 + 2(−4)x2 + 2-(−4)(2 −-4)x 4 = 1 − 4x2 + 4x4. 2! 4! 3

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  th k coefficient.

Exercise 7.2.1: Use power series methods to solve y′′ + y = 0 at the point x0 = 1 .

Exercise 7.2.2: Use power series methods to solve y′′ + 4xy = 0 at the point x0 = 0 .

Exercise 7.2.3: Use power series methods to solve y′′ − xy = 0 at the point x0 = 1 .

Exercise 7.2.4: Use power series methods to solve  ′′ 2 y + x y = 0 at the point x0 = 0 .

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 .

Exercise 7.2.6 (Chebyshev’s equation of order p ): a) Solve (1 − x2)y′′ − xy′ + p2y = 0 using power series methods at x0 = 0 . b) For what p is there a polynomial solution?

Exercise 7.2.7: Find a polynomial solution to (x2 + 1)y′′ − 2xy′ + 2y = 0 using power series methods.

Exercise 7.2.8: a) Use power series methods to solve (1 − x)y′′ + y = 0 at the point x0 = 0 . b) Use the solution to part a) to find a solution for xy′′ + y = 0 around the point x = 1 0 .

Exercise 7.2.101: Use power series methods to solve  ′′ 3 y + 2x y = 0 at the point x0 = 0 .

Exercise 7.2.102 (challenging): We can also use power series methods in nonhomogeneous equations. a) Use power series methods to solve  ′′ 1 y − xy = 1−x at the point x 0 = 0 . Hint: Recall the geometric series. b) Now solve for the initial condition y(0) = 0 , y′(0) = 0 .

Exercise 7.2.103: Attempt to solve x2y′′ − y = 0 at x 0 = 0 using the power series method of this section (x0 is a singular point). Can you find at least one solution? Can you find more than one solution?

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

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