Section 6.4 2nd-Order Equations
Now that we’ve mastered first-order LHCC equations, it’s time to explore second-order LHCC equations. These are slightly more complex but follow a similar approach. We’ll see how the characteristic equation helps us find solutions and look at the different cases that arise depending on the nature of the roots.
Consider the following second-order equation:
\begin{equation}
y'' - 5y' + 6y = 0 \tag{27}
\end{equation}
Substituting \(y = e^{rx}\) into this equation, we get:
\begin{align*}
\left[e^{rx}\right]'' - 5 \left[e^{rx}\right]' + 6 \left[e^{rx}\right] = \amp\ 0 \\
r^2 e^{rx} - 5r e^{rx} + 6 e^{rx} = \amp\ 0 \\
(r^2 - 5r + 6) e^{rx} = \amp\ 0 \qquad (e^{rx} \ne 0) \\
\text{characteristic equation} \rightarrow \qquad r^2 - 5r + 6 = \amp\ 0 \\
r_1 = 2, \quad r_2 = \amp\ 3 \text{.}
\end{align*}
Therefore, \(y = e^{2x}\) and \(y = e^{3x}\) are solutions, and the general solution is the combination of these two:
\begin{equation*}
y = C_1 e^{2x} + C_2 e^{3x} \text{.}
\end{equation*}
In general, for any second-order LHCC equation of the form:
\begin{equation}
a y'' + b y' + c y = 0 \tag{28}
\end{equation}
The characteristic equation is:
\begin{equation}
a r^2 + b r + c = 0 \tag{29}
\end{equation}
This quadratic equation will have two solutions, which could be either real or complex. The general solution depends on the nature of these solutions, and we will now explore the different cases. Before moving on, we highly recommend taking a minute to review some
concepts related to solving quadratic equations.
Let \(r_1\) and \(r_2\) be the solutions to the characteristic equation. When \(r_1 \ne r_2\text{,}\) the general solution is:
\begin{equation}
y = C_1\, e^{r_1 x} + C_2\, e^{r_2 x} \tag{30}
\end{equation}
However, if \(r_1 = r_2\text{,}\) the terms combine into like terms,
\begin{equation*}
y = C_1\, e^{r_1 x} + C_2\, e^{r_2 x} = \color{blue}\us{ \large \text{like terms}}{\us{\nwarrow}{\ul{C_1\, e^{r_1 x}}} + \us{\nearrow}{\ul{C_2\, e^{r_1 x}}}} = (\us{C}{\ub{C_1 + C_2}})\, e^{r_1 x}\text{,}
\end{equation*}
which is an incomplete solution. To fix this, we multiply the second term by
\(x\text{,}\) resulting in the correct general solution:
See
Exercise 6.6.1 for an outline of why the extra
\(x\) is required in the general solution of a second order LHCC equation when
\(r_1=r_2\text{.}\)
\begin{equation*}
y = C_1\, e^{r_1 x} + C_2\, x e^{r_1 x}
\end{equation*}
Although, the above explanation covers the situation when
\(r_1\) and
\(r_2\) are equal or not, we will split the
\(r_1 \ne r_2\) case into
two cases that treat real or complex values separately. The three cases are summarized below.
Summary of Cases.
Let \(r_1\) and \(r_2\) be the solutions of the characteristic equation associated with the second-order LHCC equation. The general solution is:
- Case 1 (real & \(\ne\))
- \(\ds y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \quad (r_1 \ne r_2)\)
- Case 2 (real & \(=\))
- \(\ds y = (C_1 + C_2 x) e^{r_1 x} \quad (r_1 = r_2)\)
- Case 3 (complex)
- \(\ds y = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x))\)
\(\ds (r = \alpha \pm i\beta \implies r_1 = \alpha + i\beta, \ r_2 = \alpha - i\beta)\)
Let’s now practice solving some second-order LHCC equations.
Example 6.
\(\ \ \)
Solution.
First, write down and solve the characteristic equation:
\begin{align*}
r^2 - 5r + 6 =\amp\ 0 \\
(r-2)(r-3) =\amp\ 0 \\
r_1 = 2 \quad \text{and} \quad r_2 =\amp\ 3 \text{.}
\end{align*}
Since
\(r_1\) and
\(r_2\) are real and
\(r_1 \ne r_2\text{,}\) we are in
Case 1. Therefore, the general solution is:
\begin{equation*}
y = C_1 e^{2x} + C_2 e^{3x} \text{.}
\end{equation*}
Solution.
The characteristic equation is:
\begin{align*}
r^2 + 4r + 4 =\amp\ 0 \\
(r+2)(r+2) =\amp\ 0 \\
r_1 = -2 \quad \text{and} \quad r_2 =\amp\ -2 \quad \leftarrow \knowl{./knowl/xref/lhcc-case-2.html}{\text{Case 2}} \text{.}
\end{align*}
Therefore, the general solution is:
\begin{equation*}
y = (C_1 + C_2 x) e^{-2x} \text{.}
\end{equation*}
Solution.
The characteristic equation in this problem is:
\begin{align*}
r^2 + 2r + 5 = 0\amp \\
r = \us{\alpha}{\boxed{-1}} \pm \us{\beta}{\boxed{2}}i \amp\ \quad \leftarrow \knowl{./knowl/xref/lhcc-case-3.html}{\text{Case 3}} \text{.}
\end{align*}
Here, \(\alpha = -1\) and \(\beta = 2\text{.}\) So the general solution is:
\begin{equation*}
y = e^{-x} (C_1 \cos(2x) + C_2 \sin(2x)) \text{.}
\end{equation*}
Reading Questions Check-Point Questions
1. Match the DE to Its Characteristic Equation.
2. Solving the characteristic equation.
If the characteristic equation for an LHCC equation is \(r^2 - 4r + 4 = 0\text{,}\) what are the roots?-
\(r = 2\) (double root)
- Correct! The root \(r = 2\) has a multiplicity of 2.
- \(r = 4\)
- Incorrect. Check the quadratic equation carefully.
-
\(r = -2\) and \(r = 2\)
- Incorrect. Ensure you solve the quadratic equation correctly.
- \(r = 2 \pm 2i\)
- Incorrect. These are not the correct roots for this equation.
3. Match the Characteristic Equation Roots to the General Solution.
Match each set of characteristic equation roots on the left to the corresponding general solution on the right.
Note: One of the differential equations is first-order, and the rest are second-order.
- \(\small r = 5 \)
- \(\small C e^{5x} \)
- \(\small r_1 = 2, \ r_2 = 3 \)
- \(\small C_1 e^{2x} + C_2 e^{3x} \)
- \(\small r_1 = -1, \ r_2 = -4 \)
- \(\small C_1 e^{-x} + C_2 e^{-4x} \)
- \(\small r_1 = 3 + 2i, \ r_2 = 3 - 2i \)
- \(\small e^{3x} (C_1 \cos(2x) + C_2 \sin(2x)) \)
- \(\small r_1 = -2, \ r_2 = -2 \)
- \(\small (C_1 + C_2 x) e^{-2x} \)
- \(\small r_1 = -1 + i, \ r_2 = -1 - i \)
- \(\small e^{-x} (C_1 \cos(x) + C_2 \sin(x)) \)
4. Discriminant of the characteristic equation.
If the discriminant of the characteristic equation is \(-4\text{,}\) then what kind of solutions does the characteristic equation have?- Real and equal solutions
- Incorrect. The discriminant is the number under the square root in the quadratic formula.
- Real and unequal solutions
- Incorrect. The discriminant is the number under the square root in the quadratic formula.
- Complex solutions
- Correct! A negative discriminant indicates the square root of a negative number in the quadratic formula, which results in the complex number \(i\text{.}\)
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