Ordinary Differential Equations/Solving Constant Coefficient Second Order Homegenous ODEs

This article concerns ODEs of the form $y''+ay'+by=0$, where $a,b$ are real constants. The basic idea with solving such ODEs is to guess a solution of the form $e^{\lambda t}$, find which values of $\lambda$ work, then write the general solution as a linear combination of these which depends on the initial conditions. Actually, this procedure works for ODEs of any order.

We will detail the procedure later in this article, but before we do so, it makes sense to step back and ask: Why the exponential function? Why do we expect this function to work and not other random ones? The most important property of the exponential function is that $e^{ct}$ is the unique solution of ${\frac {d}{dt}}y=cy$ with $y(0)=1$ (In fact, some books define the exponential function this way). We can think of ${\frac {d}{dt}}$ as something that operates on functions by mapping them to their derivative. Now, with some imagination, the differential equation ${\frac {d}{dt}}y=cy$ looks like an eigenvalue problem if we think of ${\frac {d}{dt}}$ in this manner as an operator. Then, in this analogy, the exponential function is an eigenvector (sometimes called an eigenfunction) of this operator.

We know from linear algebra that if $A$ is a matrix and $v$ is one of its eigenvectors, $v$ is an eigenvector of any power of $A$ or $A^{-1}$, and it is also an eigenvector for any linear combination of matrices for which it is an eigenvector. In particular, this means that $v$ is an eigenvector of any matrix of the form $A^{n}+c_{n-1}A^{n-1}+\cdots +c_{1}A$.

If we apply this fact to the derivative operator (analogous to a matrix), we see that the exponential functions are the eigenfunctions of any constant-coefficient differential operator and we can expect them to show up whenever we solve a constant-coefficient ODE.

Now, let's solve a second order homogenous constant coefficient ODE, $y''+ay'+by=0$. We first guess a solution of the form $e^{\lambda t}$. If we plug in our guess, we get

$(\lambda ^{2}+a\lambda +b)e^{\lambda t}=0.$

The second term in this product is never zero, so it follows that $e^{\lambda t}$ is a solution to the ODE if and only if $\lambda ^{2}+a\lambda +b=0$. $\lambda ^{2}+a\lambda +b$ is known as the characteristic equation for this ODE, and solving it will find all the values of $\lambda$ such that $e^{\lambda t}$ is a solution to the ODE. Because we are considering linear, homogenous ODEs, any linear combination of such solutions will be a solution, so in general, the solution will be one such linear combination. Initial conditions need to be specified to get a unique solution.

This is a good place to do an example. Consider the ODE

$y''-y=0.$

We guess $e^{\lambda t}$ is a solution, and find $\lambda ^{2}-1=0$, implying the two possible values for $\lambda$ are 1 and -1. This means in general, a solution to this ODE looks like

$y(t)=c_{1}e^{t}+c_{2}e^{-t}$

where $c_{1},c_{2}$ are real constants.

Suppose we specify that we want $y(0)=0,y'(0)=1$. Then, if we plug in these conditions into our general solution, we find $c_{1}=-c_{2},c_{1}=1/2$, which means

$y(t)={\frac {e^{t}-e^{-t}}{2}}.$

This particular combination of exponentials we have found has a special name. The function $y$ is called the hyperbolic sine, often written $\sinh t$ (pronounced "sinch").

Now we know how to solve ODEs of this type. But there is one thing we haven't addressed: What if the characteristic equation doesn't have two real roots? The problem when the characteristic equations has no real roots is clear, you don't have any solutions. If the characteristic equation has one repeated root, the issue is that this method only gives you one solution (up to scaling), and you don't have enough freedom to enforce two initial conditions. We will discuss these cases in another page.