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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.”

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

Let us start with a general homogeneous linear equation

(2.4) |

Theorem 2.3.1 (Superposition). Suppose , , …, are solutions of the homogeneous equation (2.4). Then

also solves (2.4) for arbitrary constants .

In other words, a linear combination of solutions to (2.4) is also a solution to (2.4). We also have the existence and uniqueness theorem for nonhomogeneous linear equations.

Theorem 2.3.2 (Existence and uniqueness). Suppose through , and are continuous functions on some interval , is a number in , and are constants. The equation

has exactly one solution deﬁned on the same interval satisfying the initial conditions

When we had two functions and we said they were linearly independent if one was not the multiple of the other. Same idea holds for functions. In this case it is easier to state as follows. The functions , , …, are linearly independent if

has only the trivial solution , where the equation must hold for all . If we can solve equation with some constants where for example , then we can solve for as a linear combination of the others. If the functions are not linearly independent, they are linearly dependent.

Example 2.3.1: Show that are linearly independent.

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.

Let us write down

We use rules of exponentials and write . Hence and . Then we have

The left hand side is a third degree polynomial in . It is either identically zero, or it has at most 3 zeros. Therefore, it is identically zero, , and the functions are linearly independent.

Let us try another way. As before we write

This equation has to hold for all . We divide through by to get

As the equation is true for all , let . After taking the limit we see that . Hence our equation becomes

Rinse, repeat!

How about yet another way. We again write

We can evaluate the equation and its derivatives at diﬀerent values of to obtain equations for , , and . Let us ﬁrst divide by for simplicity.

We set to get the equation . Now diﬀerentiate both sides

We set to get . We divide by again and diﬀerentiate to get . It is clear that is zero. Then must be zero as , and must be zero because .

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.

Example 2.3.2: On the other hand, the functions , , and are linearly dependent. Simply apply deﬁnition of the hyperbolic cosine:

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 order we need to ﬁnd linearly independent solutions. It is best seen by example.

Example 2.3.3: Find the general solution to

(2.5) |

Try: . We plug in and get

We divide through by . Then

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

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 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 into our polynomial and get ; we plug in and get 3. That means there is a root between and , because the sign changed. If we ﬁnd one root, say , then we know is a factor of our polynomial. Polynomial long division can then be used.

A good strategy is to begin with , 1, or 0. These are easy to compute. Our polynomial happens to have two such roots, and . There should be 3 roots and the last root is reasonably easy to ﬁnd. The constant term in a monic polynomial such as this is the multiple of the negations of all the roots because . So

You should check that really is a root. Hence we know that , and 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. Hence the general solution is

Suppose we were given some initial conditions , , and . Then

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 § 3.2. For now we note that the solution is , , and . The speciﬁc solution to the ODE isNext, suppose that we have real roots, but they are repeated. Let us say we have a root repeated times. In the spirit of the second order solution, and for the same reasons, we have the solutions

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

We note that the characteristic equation is

By inspection we note that . Hence the roots given with multiplicity are . Thus the general solution is

The case of complex roots is similar to second order equations. Complex roots always come in pairs . Suppose we have two such complex roots, each repeated times. The corresponding solution is

where , …, , , …, are arbitrary constants.

The characteristic equation is

Hence the roots are , both with multiplicity 2. Hence the general solution to the ODE isThe 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.

Exercise 2.3.4: Suppose the characteristic equation for a diﬀerential equation is . a) Find such a diﬀerential equation. b) Find its general solution.

Exercise 2.3.5: Suppose that a fourth order equation has a solution . a) Find such an equation. b) Find the initial conditions that the given solution satisﬁes.

Exercise 2.3.6: Find the general solution for the equation of Exercise 2.3.5.

Exercise 2.3.7: Let , , and . Are , , and linearly independent? If so, show it, if not, ﬁnd a linear combination that works.

Exercise 2.3.8: Let , , and . Are , , and linearly independent? If so, show it, if not, ﬁnd a linear combination that works.

Exercise 2.3.9: Are , , and linearly independent? If so, show it, if not, ﬁnd a linear combination that works.

Exercise 2.3.10: Are , , and linearly independent? If so, show it, if not, ﬁnd a linear combination that works.

Exercise 2.3.102: Suppose that the characteristic equation of a third order diﬀerential equation has roots 3,. a) What is the characteristic equation? b) Find the corresponding diﬀerential equation. c) Find the general solution.

Exercise 2.3.104: Are , , , linearly independent? If so, show it, if not ﬁnd a linear combination that works.