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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.

Find the general solution for each LHCC equation.
\begin{equation*} 3y' - 9y = 0 \hspace{5em} 5y' = \pi y \end{equation*}
Solution. 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.

    Select the characteristic equation for the first-order LHCC equation
    \begin{equation*} 4y' + 3y = 0\text{?} \end{equation*}
  • \(4r + 3 = 1\)
  • Incorrect. Remember, the characteristic equation should be formed by setting the equation to zero.
  • \(4r + 3 = 0\)
  • Correct! The characteristic equation is found by substituting \(y = e^{rx}\) into the differential equation.
  • \(4r - 3 = 0\)
  • Incorrect. Check the signs of the coefficients.
  • \(3r + 4 = 0\)
  • Incorrect. Ensure you have used the correct coefficients from the original equation.

3. Select the general solution.

    Give the general solution for the first-order LHCC equation
    \begin{equation*} y' - 2y = 0\text{?} \end{equation*}
  • \(\ds y = Ce^{2x}\)
  • Incorrect. Check the sign of the exponent.
  • \(\ds y = Ce^{-2x}\)
  • Correct! The solution has the form \(y = Ce^{rx}\text{,}\) where \(r\) is found from the characteristic equation.
  • \(\ds y = Ce^{\frac12 x}\)
  • Incorrect. Make sure to solve the characteristic equation correctly.
  • \(\ds y = Ce^{-\frac12 x}\)
  • Incorrect. Check the coefficient of \(y\) in the original equation.

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.

    What exponential term in the solution of the equation
    \begin{equation*} -2y' - 7y = 0\text{?} \end{equation*}
  • \(\ds e^{-7/2x}\)
  • Correct! Solving \(-2r - 7 = 0\) gives \(r = -\frac{7}{2}\text{.}\) So \(\ds e^{-7/2x}\) is the exponential term in the solution.
  • \(\ds e^{7/2x}\)
  • Incorrect. Check the signs when solving the characteristic equation.
  • \(\ds e^{-2/7x}\)
  • Incorrect. Ensure you are solving the characteristic equation correctly.
  • \(\ds e^{2/7x}\)
  • Incorrect. Revisit the steps to solve the characteristic equation.
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