Definition, Motivation of the Test Solution
We will address the easiest type of second-order constant-coefficient homogeneous ODEs first. As the name suggests, they are ODEs in the form of
\begin{equation}
\frac{d^2y}{dx^2} + \alpha\frac{dy}{dx} + \beta y = 0 \tag{1}
\end{equation}
where the coefficients \(\alpha,\beta\) are some constants and the R.H.S. is zero. Inspired by the first-order counterpart:
\begin{equation}
\frac{dy}{dx} + ky = 0 \tag{2}
\end{equation}
which can be treated by Separation of Variables and is easily seen to have the solution of
\begin{align}
\frac{dy}{dx} &= -ky \\
\int \frac{dy}{y} &= \int -kdx \\
\ln |y| &= -kx + C \\
y &= Ae^{-kx} \tag{3}
\end{align}
we are motivated to make the similar guess \( e^{rx} \) as the test solution (or called ansatz) to (1), where \( r \) is some constant to be determined. Substituting this into (1) produces
\begin{align}
\frac{d^2}{dx^2}(e^{rx}) + \alpha\frac{d}{dx}(e^{rx}) + \beta(e^{rx}) &= 0 \\
r^2e^{rx} + \alpha re^{rx} + \beta e^{rx} &= 0 \\
e^{rx}(r^2 + \alpha r + \beta) &= 0 \\
\Rightarrow r^2 + \alpha r + \beta &= 0 \tag{4}
\end{align}
because the exponential is never zero. The relation \(r^2 + \alpha r + \beta = 0\) is known as the auxiliary equation. By the derivation above, provided that this auxiliary equation is satisfied, \( e^{rx} \) will be a valid solution to (1). The remaining problem is then to compute the roots \( r_1, r_2 \) of the auxiliary equation.
Three Cases for the Auxiliary Equation
The auxiliary equation is essentially a quadratic equation, which is well-known that the behavior can be categorized into three cases from high-school algebra. These three cases will lead to correspondingly different forms of the solutions.
Two Real Roots
The first scenario is the auxiliary equation having two distinct real roots \( r_1, r_2 \), when the discriminant is greater than zero, \( \alpha^2 -4\beta > 0\). Then as aforementioned, \( y_1 = e^{r_1x} \) and \( y_2 = e^{r_2x} \) will be two solutions. They are linearly independent by computing the Wronskian (see the last tutorial):
\begin{align}
W[y_1, y_2](x) &=
\begin{vmatrix}
e^{r_1x} & e^{r_2x} \\
(e^{r_1x})’ & (e^{r_2x})’
\end{vmatrix} \\
&= \begin{vmatrix}
e^{r_1x} & e^{r_2x} \\
r_1e^{r_1x} & r_2e^{r_2x}
\end{vmatrix} \\
&= e^{r_1x}(r_2e^{r_2x}) -e^{r_2x}(r_1e^{r_1x}) \\
&= (r_2 -r_1)e^{(r_1+r_2)x} \neq 0 && \text{($r_1 \neq r_2$)} \tag{5}
\end{align}
So by linearity, the general solution will be \( y = c_1e^{r_1x} + c_2e^{r_2x} \).
Example
Solve the second-order constant-coefficient ODE
\begin{equation}
\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0 \tag{6}
\end{equation}
where the initial conditions are \( y(0) = 3\), \( y'(0) = -5 \).
The auxiliary equation is just proceeded as
\begin{align}
r^2 + 3r + 2 &= 0 \\
(r+1)(r+2) &= 0 \\
\Rightarrow r &= -1 \text{ or } {-}2 \tag{7}
\end{align}
So the general solution is \(y = c_1e^{-x} + c_2e^{-2x} \) with \(y’ = -c_1e^{-x} -2c_2e^{-2x} \). Substituting the I.C. gives
\begin{align}
&\left\{\begin{aligned}
3 &= c_1 + c_2 \\
-5 &= -c_1 -2c_2
\end{aligned}\right. \\
&\Rightarrow c_1 = 1, c_2 = 2 \\
&\Rightarrow y = e^{-x} + 2e^{-2x} \tag{8}
\end{align}
Exercise
Find the full solution for the ODE
\begin{equation}
\frac{d^2y}{dx^2} -6\frac{dy}{dx} + y = 0 \tag{9}
\end{equation}
with the initial conditions \( y(0) = 1 \), \( y'(0) = 3 \).
Answer
The auxiliary equation \( r^2 -6r + 1 = 0 \) has two real roots:
\begin{align}
r &= \frac{-(-6) \pm \sqrt{(-6)^2 – 4(1)(1)}}{2(1)} \\
&= \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}
\end{align}
So the general solution takes the form of \(y = c_1e^{(3+2\sqrt{2})x} + c_2e^{(3-2\sqrt{2})x} \). The initial conditions suggest that
\begin{align}
&\left\{\begin{aligned}
c_1 + c_2 &= 1 \\
(3+2\sqrt{2})c_1 + (3-2\sqrt{2})c_2 &= 3
\end{aligned}\right. \\
&\Rightarrow c_1 = \frac{1}{2}, c_2 = \frac{1}{2}
\end{align}
and thus \( y = \frac{1}{2}e^{(3+2\sqrt{2})x} + \frac{1}{2}e^{(3-2\sqrt{2})x} \).
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