Extension: Solving y'' + ay' + by = 0

$$0 = y'' + ay' + by$$

To solve this equation, we look at the characteristic polynomial, which for this equation is $$r^2 + ar + b = 0.$$ The reason this polynomial matters is that if we consider a solution of the form $y = e^{rx}$, and plug it in to the differential equation (in order to solve for $r$), we get $$0 = r^2e^{rx} + are^{rx}+be^{rx} = (r^2+ar+b)e^{rx}$$ so it is quite natural that the solutions of the differential equation are tied up with the roots of $r^2 + ar + b = 0.$

As with all quadratics, we have three cases:

1. The roots, $r_1$ and $r_2$, are real and different. Then the solutions are given by $$y=c_1e^{r_1x}+c_2e^{r_2x}.$$ Notice that we can add individual solutions and multiply them by constants to get the general solution only because the equation is homogenous.

2. The roots are real and the same. Then if $r$ is the common root, the solutions are given by $$y = c_1 e^{rx} + c_2 xe^{rx}.$$ The reader is encouraged to verify that this does indeed work.

3. The roots are nonreal (and therefore conjugates). They are then of the form $r_{1,2} = \lambda \pm \mu i$ for some real $\lambda$ and $\mu$. Just as in the real case, the solutions are given by $y=c_1e^{r_1x}+c_2e^{r_2x}.$ However, since the $r_{1,2}$ are nonreal, this is not a good form to leave it in; the solution to a differential equation in reals ought to be real. By Euler's Formula, we can transform these solutions into the result $$y = e^{\lambda x}\left(c_1\cos(\mu x) + c_2 \sin(\mu x)\right).$$

This is a 2nd order linear homogeneous differential equation with constant coefficients.

Let's look at those properties one at a time. Each term with $y$ or its derivatives is of the form $c(x)y^{(n)}$. (The notation $y^{(n)}$ means the nth derivative of $y$.) That makes this equation linear.

The $c(x)$ here are constants, namely 1, $a$, and $b$; hence "constant coefficients".

Furthermore, if $y(x)$ is a solution, then $k\cdot y(x)$ is also a solution; hence "homogenous."

We now know the solution for this second-order differential equation, which depends on the solutions to the characteristic polynomial $ar^2+br+c=0$. These are repeated below for convenience.

If the characteristic polynomial has two different real roots $r_1$ and $r_2$, then the solutions are $y=c_1 e^{r_1x}+c_2e^{r_2x}$, for any real $c_1,c_2$. Here's an example: $y''+5y'+6y=0$.

If the characteristic polynomial has a repeated real root $r$, then the solutions are given by $y=c_1e^{rx}+c_2xe^{rx}$, for any real $c_1,c_2$. Here's an example: $y''+6y'+9y=0$.

If the characteristic polynomial has two complex roots (which must be conjugates) given by $\lambda \pm \mu i$, then the solutions are $y=c_1 e^{\lambda x}\cos(\mu x)+c_2 e^{\lambda x} \sin(\mu x)$, for any real $c_1,c_2$. Here's an example: $y''+4y=0$.

For each of the examples given, take a moment to find the solutions described above. Plug the solutions back into the differential equation and verify that they indeed work.

The goal of this article is to answer the question: how do we know we have all of the solutions? How do we know that, in addition to the above solutions, there aren't any more?

To do this, we will be using a specific case of the ** Existence and Uniqueness** theorem, a cornerstone of differential equations. For our problem, the statement is that there exists a unique solution to the differential equation $ay''+by'+cy=0$ along with the two initial conditions $y(x_0)=y_0$ and $y'(x_0)=y_0'$. Here, $y_0$ and $y_0'$ are real constants.

By the above reasoning, we have that $y_1+y_2$ and $ay_1$, for any real $a$, are solutions to our differential equation, given that $y_1$ and $y_2$ are solutions. This is known as linearity, and arises from the fact that the derivative is a linear operator.

Extending this logic, if we have two solutions $y_1$ and $y_2$, we have a whole host of other solutions; namely, $y=c_1y_1+c_2y_2$, for real constants $c_1,c_2$.

Now, we must take the initial conditions into account. If $y=c_1y_1+c_2y_2$ satisfies the initial conditions, then

$$\begin{equation} c_1y_1(x_0)+c_2y_2(x_0)=y_0 \end{equation}$$ $$\begin{equation} c_1y_1'(x_0)+c_2y_2'(x_0)=y_0'. \end{equation}$$

There's a lot of variables here. Remember that we already have the functions $y_1$ and $y_2$, and that the initial condition parameters ($x_0$, $y_0$, $y_0'$) are given as part of the problem. We must determine the values of $c_1$ and $c_2$. One way to do this is to multiply the first equation by $y_1(x_0)$ and the second by $y_1'(x_0)$. We find that \begin{equation} c_1y_1(x_0)y_1'(x_0)+c_2y_2(x_0)y_1'(x_0)=y_0y_1'(x_0) \end{equation} \begin{equation} c_1y_1'(x_0)y_1(x_0)+c_2y_2'(x_0)y_1(x_0)=y_0'y_1(x_0). \end{equation}

Define the Wronskian $W(y_1,y_2)$ as below: $$\begin{equation} W(y_1,y_2)=y_1y_2'-y_1'y_2. \end{equation}$$ By the reasoning in the above question, if $y_1$ and $y_2$ are solutions to our differential equation, then $y=c_1y_1+c_2y_2$ can be a solution only if $W(y_1,y_2)(x_0) \neq 0$.

Here's the punchline:

Suppose we already have two solutions $y_1$ and $y_2$ to the differential equation $ay''+by'+cy=0$. If there exists a point $x_1$ for which $W(y_1,y_2)(x_1) \neq 0$,then ALL solutions to the differential equation can be written as $y=c_1y_1+c_2y_2$.

The proof uses the Existence and Uniqueness Theorem. Let $y=f(x)$ be a solution to the differential equation, which we know must exist. Then, since the Wronskian is nonzero at at least one point $x_0$, we can solve the equations \begin{equation} c_1y_1(x_0)+c_2y_2(x_0)=f(x_0) \end{equation} \begin{equation} c_1y_1'(x_0)+c_2y_2'(x_0)=f'(x_0) \end{equation} for $c_1$ and $c_2$. Thus, $y=c_1y_1+c_2y_2$ is a solution, but by Uniqueness, we must have that $f(x)=c_1y_1+c_2y_2$, as desired!

Intuitively, the Wronskian tells us whether or not the solutions $y_1$ and $y_2$ carry different information. If they are really the same solution, then we would need more information before finding a general solution by $y=c_1y_1+c_2y_2$. Furthermore, two solutions $y_1$ and $y_2$ are called fundamental solutions if their Wronskian is nonzero over an interval.

Let's apply our knowledge to the case of repeated roots now. As you have checked above, both $y_1=e^{rx}$ and $y_2=xe^{rx}$ are solutions. Before concluding that all solutions are given by $y=c_1e^{rx}+c_2xe^{rx}$, we must check that $y_1$ and $y_2$ are a fundamental set of solutions. Note that \begin{equation} W(y_1,y_2)=e^{rx}(rxe^{rx}+e^{rx})-xe^{rx}(re^{rx})=e^{2rx} \neq 0, \end{equation} so indeed all solutions are given by $c_1y_1+c_2y_2$.