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Section 9.3 Second-Order Equations
Second-order linear homogeneous constant coefficient (LHCC) equations appear in many applications across physics, engineering, and applied mathematics. These equations have the general form:
\begin{equation}
a y'' + b y' + c y = 0\tag{9.4}
\end{equation}
where \(a\text{,}\) \(b\text{,}\) and \(c\) are constants. In this section, we’ll explore how the solutions depend on the roots of the associated characteristic equation, and how different types of roots give rise to different forms of the general solution.
Subsection Fundamental Solutions
Checkpoint 132 . 📖❓ Classifying Characteristic Equations.
To solve
(9.4) , we begin by forming the characteristic equation as we did in the previous section. For this case, the characteristic equation is the
quadratic
\begin{equation}
a r^2 + b r + c = 0\tag{9.5}
\end{equation}
\begin{equation}
r_1 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}, \quad r_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\text{.}\tag{9.6}
\end{equation}
Each root corresponds to an exponential solution—either
\(e^{r_1 x}\) or
\(e^{r_2 x}\text{.}\) These are called the
fundamental solutions of the differential equation.
Subsection The General Solution
Once we have the fundamental solutions, the superposition principle tells us that
\begin{equation*}
y = c_1 e^{r_1 x} + c_2 e^{r_2 x}
\end{equation*}
is a solution to
(9.4) for any constants
\(c_1\) and
\(c_2\text{.}\)
However, if the roots
\(r_1\) and
\(r_2\) are equal, then
\(e^{r_1 x}\) and
\(e^{r_2 x}\) are like-terms. In that case, the expression above collapses into a single term, and the solution is incomplete.
To create a second, independent solution, we multiply the second term by \(x\text{,}\) giving:
\begin{equation*}
y = c_1 e^{r x} + c_2 x e^{r x}\text{.}
\end{equation*}
This guarantees a complete general solution with independent terms. There are three distinct scenarios to consider, based on the discriminant of the characteristic equation:
\begin{equation*}
\Delta = b^2 - 4ac\text{.}
\end{equation*}
Each case leads to a different solution form, broken down as follows.
Subsection Case 1 \((\Delta > 0)\text{:}\) Real, Non-Repeating Solutions
If \(\Delta\) is positive, then the characteristic equation has the two unequal real solutions:
\begin{equation*}
r_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad
r_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a},
\end{equation*}
so the general solution is
\begin{equation*}
y = c_1 e^{r_1 x} + c_2 e^{r_2 x}.
\end{equation*}
Subsection Case 2 \((\Delta < 0)\text{:}\) Two Complex Solutions
If \(\Delta\) is negative, the characteristic equation has two complex conjugate solutions:
\begin{equation*}
r_1 = \alpha + i \beta, \qquad r_2 = \alpha - i \beta\text{.}
\end{equation*}
The corresponding general solution is:
\begin{equation*}
y = c_1 e^{(\alpha + i \beta)x} + c_2 e^{(\alpha - i \beta)x}\text{.}
\end{equation*}
However, it is common to express this using sine and cosine (via
Euler’s Formula ):
\begin{equation*}
y = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\text{.}
\end{equation*}
Subsection Case 3 \((\Delta = 0)\text{:}\) Real, Repeating Solutions
If \(\Delta\) is zero, then the characteristic equation has a repeated root:
\begin{equation*}
r_1 = r_2 = \frac{-b}{2a}\text{.}
\end{equation*}
In this case, the general solution is:
\begin{equation*}
y = c_1 e^{r x} + c_2 x e^{r x}\text{.}
\end{equation*}
Subsection Summary & Examples
The form of the general solution to a second-order LHCC equation depends entirely on the roots of its characteristic equation. Here’s a summary of the three cases.
✳️ 2nd-Order LHCC General Solution Cases.
Case 1 \((\Delta > 0)\) \(\small \text{unequal real roots}\)
\(\ds y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)
Case 2 \((\Delta < 0)\) \(\small \text{complex roots}\)
\(r = \alpha \pm i\beta\)
\(\ds y = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\)
Case 3 \((\Delta = 0)\) \(\small \text{repeated real root}\)
\(\ds y = c_1 e^{r x} + c_2 x e^{r x}\)
Checkpoint 133 . 📖❓ Match the Characteristic & General Solutions.
Match the characteristic solutions to the corresponding general solution.
\(\small r = 5 \) \(\small C e^{5x} \) \(\small r_1 = 2, \ r_2 = 3 \) \(\small C_1 e^{2x} + C_2 e^{3x} \) \(\small r_1 = -1, \ r_2 = -4 \) \(\small C_1 e^{-x} + C_2 e^{-4x} \) \(\small r_1 = 3 + 2i, \ r_2 = 3 - 2i \) \(\small e^{3x} (C_1 \cos(2x) + C_2 \sin(2x)) \) \(\small r_1 = -2, \ r_2 = -2 \) \(\small (C_1 + C_2 x) e^{-2x} \) \(\small r_1 = -1 + i, \ r_2 = -1 - i \) \(\small e^{-x} (C_1 \cos(x) + C_2 \sin(x)) \)
🌌 Example 134 . Case 1 Examples.
Find the general solution to each second-order LHCC equation.
\(\ds y'' - 5y' + 6y = 0\) Solution .
The characteristic equation is:
\begin{equation*}
r^2 - 5r + 6 = 0
\end{equation*}
Factoring gives:
\begin{equation*}
(r - 2)(r - 3) = 0
\end{equation*}
So the roots are \(r_1 = 2\) and \(r_2 = 3\text{,}\) giving the general solution:
\begin{equation*}
y = c_1 e^{2x} + c_2 e^{3x}
\end{equation*}
\(\ds 5y'' + 13y' - 2y = 0\) Solution .
The characteristic equation is:
\begin{equation*}
5r^2 + 13r - 2 = 0
\end{equation*}
Using the quadratic formula:
\begin{align*}
r \amp = \frac{-13 \pm \sqrt{13^2 - 4(5)(-2)}}{2 \cdot 5}\\
\amp = \frac{-13 \pm \sqrt{201}}{10}
\end{align*}
Since the roots are unequal and real, the general solution is:
\begin{equation*}
y = c_1 e^{r_1 x} + c_2 e^{r_2 x}
\end{equation*}
where
\(r_1 = \frac{-13 - \sqrt{201}}{10}\) and
\(r_2 = \frac{-13 + \sqrt{201}}{10}\text{.}\)
🌌 Example 135 . Case 2 Examples.
Find the general solution to each second-order LHCC equation.
\(\ds \omega'' + 4\omega' + 5\omega = 0\) Solution .
The characteristic equation is:
\begin{equation*}
r^2 + 4r + 5 = 0
\end{equation*}
Applying the quadratic formula:
\begin{align*}
r \amp = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}\\
\amp = -2 \pm i
\end{align*}
So \(\alpha = -2\text{,}\) \(\beta = 1\text{,}\) and the general solution is:
\begin{equation*}
\omega(t) = e^{-2t}(c_1 \cos t + c_2 \sin t)
\end{equation*}
\(\ds y'' + 25y = 0\) Solution .
The characteristic equation is:
\begin{equation*}
r^2 + 25 = 0
\end{equation*}
This gives \(r = \pm 5i\text{,}\) so \(\alpha = 0\) and \(\beta = 5\text{.}\) The general solution is:
\begin{equation*}
y = e^{0x}\left(c_1\cos(5x) + c_2\sin(5x)\right)
\end{equation*}
or simply:
\begin{equation*}
y = c_1 \cos(5x) + c_2 \sin(5x)
\end{equation*}
🌌 Example 136 . Case 3 Examples.
Find the general solution to each second-order LHCC equation.
\(\ds y'' + 4y' + 4y = 0\) Solution .
The characteristic equation is:
\begin{equation*}
r^2 + 4r + 4 = 0
\end{equation*}
Factoring gives:
\begin{equation*}
(r + 2)^2 = 0
\end{equation*}
So \(r = -2\) is a repeated root, and the general solution is:
\begin{equation*}
y = c_1 e^{-2x} + c_2 x e^{-2x}
\end{equation*}
\(\ds 16\frac{d^2q}{dt^2} - 8\frac{dq}{dt} + q = 0\) Solution .
The characteristic equation is:
\begin{equation*}
16r^2 - 8r + 1 = 0
\end{equation*}
Using the quadratic formula:
\begin{equation*}
r \amp = \frac{8 \pm \sqrt{64 - 64}}{32} = \frac{8}{32} = \frac{1}{4}
\end{equation*}
This is a repeated root, so the general solution is:
\begin{equation*}
q(t) = c_1 e^{\frac{1}{4}t} + c_2 t e^{\frac{1}{4}t}
\end{equation*}
These examples illustrate how second-order LHCC equations are solved using the
standard ,
repeating , and
complex cases discussed above. In the next section, we will see that a similar process can be generalized to higher-order LHCC equations.
Checkpoint 137 . 🤔💭 Reading Questions.
(a) 🤔💭Solving the characteristic equation.
If the characteristic equation for an LHCC equation is
\(r^2 - 4r + 4 = 0\text{,}\) what is the general solution?
\(\quad y = c_1 e^{2x} + c_2 x e^{2x}\)
Correct! The solution \(r = 2\) has a multiplicity of 2.
\(\quad y = c_1 e^{2x} + c_2 e^{2x}\)
Incorrect. The solution \(r = 2\) has a multiplicity of 2.
\(\quad y = c_1 e^{2x} + c_2 e^{-2x}\)
Incorrect. Ensure you solve the characteristic equation correctly.
\(\quad y = e^{2x}(c_1 \cos(2x) + c_2 \sin(2x))\)
Incorrect. The characteristic equation does not have complex solutions.
(b) 🤔💭Purely Imaginary Characteristic Solutions.
If a characteristic equation has roots \(\pm i\text{,}\) the general solution is
\begin{equation*}
y = c_1 \cos x + c_2 \sin x\text{.}
\end{equation*}
True.
Correct! The complex exponentials transform into sine and cosine terms.
False.
Correct! The complex exponentials transform into sine and cosine terms.
(c) 🤔💭Matching Equations to Solutions.
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