The full description of these equations is: Linear constant coefficient homogeneous equations. The equations described in the title have the form

Here y is a function of x, and , ... , are {\it constants}. *Linear* means the
equation is a sum of the derivatives of y, each multiplied by x
stuff. (In this case, the x stuff is constant.) *Homogeneous*
means that the right side is 0 --- there's no term involving only x.

It's convenient to let stand for the
operation of differentiating with respect to x. (Note that is the *operation* of
differentiation, whereas is the
derivative.) In this notation, computes the second
derivative, computes the third derivative, and
so on. The equation above becomes

* Example.* The following equations are linear
homogeneous equations with constant coefficients:

A * solution* to the equation is a function which satisfies the equation. Equivalently,
if you think of as
a linear transformation, it is an element of the kernel of the
transformation.

The * general solution* is a linear combination
of the elements of a basis for the kernel, with the coefficients
being arbitrary constants.

The form of the equation makes it reasonable that a solution should be a function whose derivatives are constant multiples of itself. is such a function:

Plug into

The result:

Factor out and cancel it. This leaves

Thus, is a solution to the original equation
exactly when m is a root of this polynomial. The polynomial is called
the * characteristic polynomial*; as the
derivation showed, it's obtained by building a polynomial using the
coefficients of the original differential equation.

* Example.* Solve .

The characteristic polynomial is ; solving yields , so or . The general solution is

You can check this by plugging back in. Here are the derivatives:

Therefore,

* Example.* Solve .

It's easy to write down the characteristic equation: just replace the D's with m's:

The roots are and . (Don't fall into the trap of assuming that roots must be integers, or even rationals!) The solution is

What happens if there are repeated roots? Look at the equation . The characteristic equation is , which has as a double root. It is true that is a solution, but it would be incorrect to write .

The terms and are redundant --- you could combine them to get . To put it another way, as
function and are *linearly
dependent*.

It is reasonable to suppose that for a second order equation you
should have two *different* solutions. is one; how can you find another?

The idea is to guess the form that such a solution might take. Guess:

i.e. something times the known solution . What should f be?

To find f, plug into the equation. Here are the derivatives:

Plug them in:

Hence, . Integrate twice and obtain . Thus,

In fact, this is the general solution --- notice the two arbitrary constants. The functions and are indpendent solutions to the original equation.

In general, if m is a repeated root of multiplicity k in the characteristic polynomial, you get terms , , ... , in the general solution.

* Example.* Solve .

The characteristic equation is , which has as a root with multiplicity 3. The general solution is

* Example.* Solve .

The characteristic equation is , or . The roots are 1, 2, and -1 (double). The general solution is

Note: You can write the terms in the solution in any order you please. Nor does it matter which "c" goes with which term, since they {\it are} arbitrary constants.

* Example.* (* Linear
systems*) Suppose x and y are functions of t. Consider the system
of differential equations

I want to solve for x and y in terms of t.

Solve the second equation for x:

Differentiate:

Plug the expressions for x and into the first equation:

Simplify:

The characteristic equation is , or . The roots are and . Therefore,

Now , so

There are other ways of solving linear systems, but for small systems brute force works reasonably well!

Now suppose the characteristic equation has a complex root . From basic algebra, complex roots of real
polynomials come in * conjugate pairs*: and . It's reasonable to expect
solutions

However, these are complex solutions, and you should have real
solutions to the original real differential equation. I'll use the
* complex exponential formula*

You can derive this formula by considering the Taylor series for , , and .

Now

Let and . Observe that and can be solved for in terms of and , so no generality is lost with this substitution. Then

Each pair of conjugate complex roots in the characteristic equation generates a pair of independent solutions of this form.

* Example.* Solve .

The characteristic equation has roots . The solution is

* Example.* Solve .

The characteristic equation has roots . The solution is

* Example.* Solve .

The characteristic polynomial factors into . The roots are and . The solution is

* Example.* Solve .

The characteristic equation has repeated complex roots: (each double). The solution is

(What's the solution to ?)

Copyright 2008 by Bruce Ikenaga