Homogeneous Linear ODE Basics (Part 1)

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Introduction and Definitions

In order to prepare for a later article about critically damped springs, let's first go over the basics of what are called ordinary differential equations, or ODEs for short. We're going to talk specifically about homogeneous linear ODEs.

When you think of solving equations, you may think of things like 0 = x² + x, where you have an equation that involves a linear combination of powers of a particular variable (linear meaning that they're multiplied by simple scalars and added together). Let's consider another type of equation where we consider x(t) to be a function of another variable t. In this case we call x a dependent variable and t is an independent variable. Let's take a linear combination of derivatives of x with respect to t, e.g.



This is an example of a homogeneous linear ordinary differential equation. "Homogeneous" means that there are only terms that involve x or one of its derivatives; there is nothing that involves the independent variable t directly, or any constant terms, etc. "Linear" means that it's a linear combination of derivatives of x. "Ordinary" refers to the fact that there are no partial derivatives (since there's only one independent variable). In general it takes the form



Where L(x) is an operator we define on the dependent variable x.

We should note that the sum of two solutions is also a solution. Let's say we have two different solutions to (1) that we'll call x₀ and x₁. If we substitute x₀ + x₁ into (1)



The last step is due to the fact that x₀ and x₁ satisfy L(x) = 0, and so x₀ + x₁ is a solution.

Building Solutions

Before we try to find a general solution to this, let's look at a specific case



We know of one function that is equal to its derivative, which is x = e^t. We can introduce a scalar to make it x = ce^t and this is still a solution of (2). If we were to scale the exponent by a constant r, we would have



which isn't a solution to (2), but it is a solution to



where we define this as an operator F_r(x) (we'll refer back to this later on).

Now let's go back to the general equation (1) and assume that it has solutions of the form x = c e^rt. Substituting into (1) we get



We can divide out c e^rt to give us what's called the characteristic equation of our original ODE



which is just a polynomial in r for which we need to find the roots.

Distinct Roots

First let's assume that all the roots are distinct. If we factor the characteristic equation we would have



Since each of these distinct roots yields a solution of the form x = c_i e^r_it and using the fact that a sum of these solutions is a solution itself, we have the general solution

Repeated Roots

It's important to note that if we have some repeated roots, the factored characteristic equation will not look like it does in (5). For example 0 = r² - 4 r + 4 would be 0 = (r - 2)(r - 2) or 0 = (r - 2)². Since 2 is the only root of this equation, we only end up with one term in our general solution (6), so we need to examine repeated roots a bit more thoroughly. Let's rewrite our factorization to account for repeated roots



where there are k distinct roots and the i^th root has a multiplicity of m_i (i.e. it's repeated m_i times).

Here we'll come up with a way of "factoring" our general ODE form (1) similar to how we can factor the characteristic equation in (7).

If we compose the operators F_a and F_b (defined in (3)), we get



And if we multiply the two terms (r - a)(r - b) we get



which is the characteristic equation for (8), so we can see that we can build the operator L(x) by composing the operators F_ri as follows



So the "factor" of our general equation (1) that cooresponds to (r - r_i)^m_i is F_ri^m_i(x) which is the operator F_ri composed on itself m_i times.

The nice thing about this "factorization" is that if you have a solution x that satisfies some of the terms, it's also a solution to the general equation. If we assume x satisfies



then continuing to compose other factors will give



And so that particular x will continue to be a solution for the rest of the compositions, and so of course for the general equation itself.

Now we make a guess that the solution for a repeated root will take the form u_i(t) e^r_it, where we've replaced the constant c_i with a function of t that we call u_i(t).
We already know that u_i(t) is a constant when m_i = 1, i.e. when that root isn't repeated. If we expand the operator F_ri^m_i, we get



and in general



Then if we set this part of the general equation (1) to zero, we can divide out the exponential to get



which is only true if u_i(t) is a polynomial in t of degree m_i - 1, namely



And we can see if m_i = 1 then u_i(t) is a constant which is consistent with our general solution with distinct roots (6).

Conclusion

Accounting for both distinct and repeated roots, we can see that the general solution to (1) is



where there are k distinct roots, and the root r_i has multiplicity m_i.

We'll close with an example, and next time we'll look more in depth at the cases where the characteristic equation has complex roots.

Example

To find the general solution to 0 = x''' - 5x'' + 3x' + 9x, we first compute the characteristic equation 0 = r³ - 5r² + 3r + 9 and then factor it into 0 = (r + 1)(r - 3)(r - 3) with roots -1 (multiplicity 1) and 3 (multiplicity 2). So the general solution is