DIFFYQS Higher order linear ODEs

Section 2.3 Higher order linear ODEs

Note: somewhat more than 1 lecture, §3.2 and §3.3 in [EP], §4.1 and §4.2 in [BD]

We briefly study higher order equations. Equations appearing in applications tend to be second order. Higher order equations do appear from time to time, but generally the world around us is “second order.”

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The basic results about linear ODEs of higher order are essentially the same as for second order equations, with 2 replaced by

n .

The important concept of linear independence is somewhat more complicated when more than two functions are involved. For higher order constant coefficient ODEs, the methods developed are also somewhat harder to apply, but we will not dwell on these complications. It is also possible to use the methods for systems of linear equations from Chapter 3 to solve higher order constant coefficient equations.

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Let us start with a general homogeneous linear equation

\begin{matrix} (2.4) & y^{(n)} + p_{n - 1} (x) y^{(n - 1)} + \dots + p_{1} (x) y^{'} + p_{0} (x) y = 0 . \end{matrix}

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Theorem 2.3.1. Superposition.

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y_{1},

y_{2},

...,

y_{n}

are solutions of the homogeneous equation (2.4), then

y (x) = C_{1} y_{1} (x) + C_{2} y_{2} (x) + \dots + C_{n} y_{n} (x)

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also solves (2.4) for arbitrary constants

C_{1}, C_{2}, \dots, C_{n} .

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That is, a linear combination of solutions to (2.4) is a solution to (2.4). There is also the existence and uniqueness theorem for linear equations, including nonhomogeneous ones.

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Theorem 2.3.2. Existence and uniqueness.

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Suppose

p_{0}, p_{1}, \dots, p_{n - 1},

and

f

are continuous functions on some interval

I,

a

is a number in

I,

and

b_{0}, b_{1}, \dots, b_{n - 1}

are constants. Then the equation

y^{(n)} + p_{n - 1} (x) y^{(n - 1)} + \dots + p_{1} (x) y^{'} + p_{0} (x) y = f (x)

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has exactly one solution

y (x)

defined on the same interval

I

satisfying the initial conditions

y (a) = b_{0}, y^{'} (a) = b_{1}, \dots, y^{(n - 1)} (a) = b_{n - 1} .

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Subsection 2.3.1 Linear independence

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When we had two functions

y_{1}

and

y_{2},

we said they were linearly independent if one was not a multiple of the other. Same idea holds for

n

functions, although in this case it is easier to state as follows. The functions

y_{1},

y_{2},

...,

y_{n}

are linearly independent if the equation

c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n} = 0

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has only the trivial solution

c_{1} = c_{2} = \dots = c_{n} = 0,

where the equation must hold for all

x .

If we can solve the equation with some constants

c_{1}, c_{2}, \dots, c_{n},

where for example

c_{1} \neq 0,

then we can solve for

y_{1}

as a linear combination of the others. If the functions are not linearly independent, they are linearly dependent.

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Example 2.3.1.

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Verify that

e^{x}, e^{2 x}, e^{3 x}

are linearly independent.

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Let us give several ways to show this fact. Many textbooks (including [EP] and [F]) introduce Wronskians, but it is difficult to see why they work and they are not really necessary here.

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Consider

c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x} = 0.

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We use rules of exponentials and write

z = e^{x} .

Hence

z^{2} = e^{2 x}

and

z^{3} = e^{3 x} .

Then we have

c_{1} z + c_{2} z^{2} + c_{3} z^{3} = 0.

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The left-hand side is a third degree polynomial in

z .

It is either identically zero, or it has at most 3 zeros. Therefore, it is identically zero,

c_{1} = c_{2} = c_{3} = 0,

and the functions are linearly independent.

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Let us try another way. As before we write

c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x} = 0.

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This equation has to hold for all

x .

We divide through by

e^{3 x}

to get

c_{1} e^{- 2 x} + c_{2} e^{- x} + c_{3} = 0.

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As the equation is true for all

x,

let

x \to \infty .

After taking the limit we see that

c_{3} = 0 .

Hence our equation becomes

c_{1} e^{x} + c_{2} e^{2 x} = 0.

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Rinse, repeat!

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How about yet another way. We again write

c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x} = 0.

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We can evaluate the equation and its derivatives at different values of

x

to obtain equations for

c_{1},

c_{2},

and

c_{3} .

Let us first divide by

e^{x}

for simplicity.

c_{1} + c_{2} e^{x} + c_{3} e^{2 x} = 0.

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We set

x = 0

to get the equation

c_{1} + c_{2} + c_{3} = 0 .

Now differentiate both sides

c_{2} e^{x} + 2 c_{3} e^{2 x} = 0 .

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We set

x = 0

to get

c_{2} + 2 c_{3} = 0 .

We divide by

e^{x}

again and differentiate to get

2 c_{3} e^{x} = 0 .

It is clear that

c_{3}

is zero. Then

c_{2}

must be zero as

c_{2} = - 2 c_{3},

and

c_{1}

must be zero because

c_{1} + c_{2} + c_{3} = 0 .

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There is no one best way to do it. All of these methods are perfectly valid. The important thing is to understand why the functions are linearly independent.

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Example 2.3.2.

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On the other hand, the functions

e^{x},

e^{- x},

and

\cosh x

are linearly dependent. Simply apply definition of the hyperbolic cosine:

\cosh x = \frac{e^{x} + e^{- x}}{2} or 2 \cosh x - e^{x} - e^{- x} = 0.

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Once we have enough linearly independent solutions, we have the general solution to the homogeneous equation, just as we did for second order equations.

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Theorem 2.3.3.

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y_{1},

y_{2},

...,

y_{n}

are linearly independent solutions of the homogeneous equation (2.4), then the general solution to (2.4) can be written as

y (x) = C_{1} y_{1} (x) + C_{2} y_{2} (x) + \dots + C_{n} y_{n} (x) .

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Subsection 2.3.2 Constant coefficient higher order ODEs

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When we have a higher order constant coefficient homogeneous linear equation, the song and dance is exactly the same as it was for second order. We just need to find more solutions. If the equation is

n^{th}

order, we need to find

n

linearly independent solutions. It is best seen by example.

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Example 2.3.3.

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Find the general solution to

\begin{matrix} (2.5) & y^{‴} - 3 y^{″} - y^{'} + 3 y = 0 . \end{matrix}

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Try:

y = e^{r x} .

We plug in and get

\underset{y^{‴}}{\underset{⏟}{r^{3} e^{r x}}} - 3 \underset{y^{″}}{\underset{⏟}{r^{2} e^{r x}}} - \underset{y^{'}}{\underset{⏟}{r e^{r x}}} + 3 \underset{y}{\underset{⏟}{e^{r x}}} = 0 .

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We divide through by

e^{r x} .

Then

r^{3} - 3 r^{2} - r + 3 = 0 .

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The trick now is to find the roots. There is a formula for the roots of degree 3 and 4 polynomials but it is very complicated. There is no formula for higher degree polynomials. That does not mean that the roots do not exist. There are always

n

roots for an

n^{th}

degree polynomial. They may be repeated and they may be complex. Computers are pretty good at finding roots approximately for reasonable size polynomials.

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A good place to start is to plot the polynomial and check where it is zero. We can also simply try plugging in. We just start plugging in numbers

r = - 2, - 1, 0, 1, 2, \dots

and see if we get a hit (we can also try complex numbers). Even if we do not get a hit, we may get an indication of where the root is. For example, we plug

r = - 2

into our polynomial and get

- 15;

we plug in

r = 0

and get

3 .

That means there is a root between

r = - 2

and

r = 0,

because the sign changed. If we find one root, say

r_{1},

then we know

(r - r_{1})

is a factor of our polynomial. Polynomial long division can then be used.

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A good strategy is to begin with

r = 0,

1,

- 1 .

These are easy to compute. Our polynomial has two such roots,

r_{1} = - 1

and

r_{2} = 1 .

There should be 3 roots and the last root is reasonably easy to find. The constant term in a monic

The word monic means that the coefficient of the top degree

r^{d},

in our case

r^{3},

1 .

polynomial such as this is the multiple of the negations of all the roots because

r^{3} - 3 r^{2} - r + 3 = (r - r_{1}) (r - r_{2}) (r - r_{3}) .

3 = (- r_{1}) (- r_{2}) (- r_{3}) = (1) (- 1) (- r_{3}) = r_{3} .

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You should check that

r_{3} = 3

really is a root. Hence

e^{- x},

e^{x}

and

e^{3 x}

are solutions to (2.5). They are linearly independent as can easily be checked, and there are 3 of them, which happens to be exactly the number we need. So the general solution is

y = C_{1} e^{- x} + C_{2} e^{x} + C_{3} e^{3 x} .

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Suppose we were given some initial conditions

y (0) = 1,

y^{'} (0) = 2,

and

y^{″} (0) = 3 .

Then

\begin{aligned} 1 = y (0) & = C_{1} + C_{2} + C_{3}, \\ 2 = y^{'} (0) & = - C_{1} + C_{2} + 3 C_{3}, \\ 3 = y^{″} (0) & = C_{1} + C_{2} + 9 C_{3} . \end{aligned}

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It is possible to find the solution by high school algebra, but it would be a pain. The sensible way to solve a system of equations such as this is to use matrix algebra, see Section 3.2 or Appendix A. For now we note that the solution is

C_{1} = -^{1} /_{4},

C_{2} = 1,

and

C_{3} =^{1} /_{4} .

The specific solution to the ODE is

y = \frac{- 1}{4} e^{- x} + e^{x} + \frac{1}{4} e^{3 x} .

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Next, suppose that we have real roots, but they are repeated. Let us say we have a root

r

repeated

k

times. In the spirit of the second order solution, and for the same reasons, we have the solutions

e^{r x}, x e^{r x}, x^{2} e^{r x}, \dots, x^{k - 1} e^{r x} .

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We take a linear combination of these solutions to find the general solution.

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Example 2.3.4.

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Solve

y^{(4)} - 3 y^{‴} + 3 y^{″} - y^{'} = 0 .

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We note that the characteristic equation is

r^{4} - 3 r^{3} + 3 r^{2} - r = 0 .

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By inspection, we note that

r^{4} - 3 r^{3} + 3 r^{2} - r = r {(r - 1)}^{3} .

Hence the roots given with multiplicity are

r = 0, 1, 1, 1 .

Thus the general solution is

y = \underset{terms coming from r = 1}{\underset{⏟}{(C_{1} + C_{2} x + C_{3} x^{2}) e^{x}}} + \underset{from r = 0}{\underset{⏟}{C_{4}}} .

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The case of complex roots is similar to second order equations. Complex roots always come in pairs

r = α \pm i β .

Suppose we have two such complex roots, each repeated

k

times. The corresponding solution is

(C_{0} + C_{1} x + \dots + C_{k - 1} x^{k - 1}) e^{α x} \cos (β x) + (D_{0} + D_{1} x + \dots + D_{k - 1} x^{k - 1}) e^{α x} \sin (β x),

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where

C_{0},

...,

C_{k - 1},

D_{0},

...,

D_{k - 1}

are arbitrary constants.

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Example 2.3.5.

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Solve

y^{(4)} - 4 y^{‴} + 8 y^{″} - 8 y^{'} + 4 y = 0 .

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The characteristic equation is

\begin{aligned} r^{4} - 4 r^{3} + 8 r^{2} - 8 r + 4 & = 0, \\ {(r^{2} - 2 r + 2)}^{2} & = 0, \\ {({(r - 1)}^{2} + 1)}^{2} & = 0 . \end{aligned}

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Hence the roots are

1 \pm i,

both with multiplicity 2. Hence the general solution to the ODE is

y = (C_{1} + C_{2} x) e^{x} \cos x + (C_{3} + C_{4} x) e^{x} \sin x .

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The way we solved the characteristic equation above is really by guessing or by inspection. It is not so easy in general. We could also have asked a computer or an advanced calculator for the roots.

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