Section 6.3 1st-Order Equations
In the previous section, we learned that solutions to Linear Homogeneous Constant Coefficient (LHCC) equations often involve terms of the form \(e^{rx}\text{.}\) However, not every value of \(r\) will work. The correct values of \(r\) arise from solving an important algebraic equation, obtained by substituting \(y = e^{rx}\) into the differential equation. In this section, we will focus on this process for first-order equations before building up to higher-order cases.
Let’s start with a simple first-order equation:
\begin{equation*}
y' - 5y = 0
\end{equation*}
We want to find the value of \(r\) such that \(y = e^{rx}\) is a solution. Substituting \(y = e^{rx}\) into the differential equation gives:
\begin{align*}
\left[e^{rx}\right]' - 5 \left[e^{rx}\right] =\amp 0 \\
r e^{rx} - 5 e^{rx} =\amp 0 \\
(r - 5) e^{rx} =\amp 0
\end{align*}
Since \(e^{rx}\) is never zero, we must have:
\begin{gather*}
r - 5 = 0 \quad \Rightarrow \quad r = 5
\end{gather*}
This tells us that \(y = e^{5x}\) is a solution. Therefore, the general solution is:
\begin{equation*}
y = C e^{5x}
\end{equation*}
The equation \(r - 5 = 0\text{,}\) which gave us the value of \(r\text{,}\) is called the characteristic equation. Every LHCC equation has one. The characteristic equation gives us the values of \(r\) that we need to construct the general solution.
Applying the same approach to the general first-order LHCC equation gives us:
\begin{align*}
y=e^{rx} \quad\rightarrow\quad
ay' + by = 0 \quad\Rightarrow\quad a \left[e^{rx}\right]' + b \left[e^{rx}\right]
=\amp 0 \\
a r e^{rx} + b e^{rx} =\amp 0 \\
(ar + b) e^{rx} =\amp 0
\end{align*}
This results in the characteristic equation and general solution:
\begin{equation*}
ar + b = 0 \quad \Rightarrow \quad r = -\frac{b}{a} \quad \Rightarrow \quad y = C e^{-\frac{b}{a}x}
\end{equation*}
Now, let’s apply this method to a couple of examples.
Example 5.
\(\ \ \)
\begin{equation*}
3y' - 9y = 0 \hspace{5em} 5y' = \pi y
\end{equation*}
Solution.
First, rewrite the equations in standard form:
\begin{equation*}
y' - 3y = 0 \hspace{5em} 5y' - \pi y = 0
\end{equation*}
Now solve the characteristic equations:
\begin{equation*}
3r - 9 = 0 \quad \Rightarrow \quad r = 3 \hspace{5em} 5r - \pi = 0 \quad \Rightarrow \quad r = \frac{\pi}{5}
\end{equation*}
Thus, the general solutions are:
\begin{equation*}
y = C e^{3x} \hspace{5em} y = C e^{\frac{\pi}{5}x}
\end{equation*}
This straightforward method works for any first-order LHCC equation by using the characteristic equation. In the next sections, we’ll extend this technique to higher-order equations.
Reading Questions Check-Point Questions
1. Characteristic equation for first-order LHCC.
What is the characteristic equation for \(y' - 5y = 0\text{?}\)- \(r - 5 = 0\)
- Correct! The characteristic equation is \(r - 5 = 0\text{.}\)
- \(r + 5 = 0\)
- Incorrect. Check the sign of the coefficient of \(y\text{.}\)
- \(r^2 - 5 = 0\)
- Incorrect. The characteristic equation for a first-order LHCC is linear, not quadratic.
- \(5r - 1 = 0\)
- Incorrect. Make sure to use the correct coefficients from the original equation.
2. Select the characteristic equation.
3. Select the general solution.
4. Identify the first-order LHCC equation.
Which of the following is a first-order LHCC equation?- \(y'' + y' - y = 0\)
- Incorrect. This is a second-order equation.
- \(3y' + 5y = 0\)
- Correct! This is a first-order linear homogeneous equation with constant coefficients.
- \(2y + y' = 3\)
- Incorrect. This equation is not homogeneous.
- \(y' + xy = 0\)
- Incorrect. This is not a constant coefficient equation.
5. What exponential term is in the solution.
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