# Linear Homogeneous Differential Equations

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 ?)