DE-BOOK Second-Order Equations

Section 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:

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\begin{equation} a y'' + b y' + c y = 0\tag{34} \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.

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Checkpoint 139. 📖❓ Classifying Characteristic Equations.

Review quadratic equations and answer the following question.

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Match the type of roots with the corresponding quadratic equation.

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Good work! The discriminant helps classify root types.

Real unequal roots
\(r^2 - 4r + 3 = 0\)
Real repeated roots
\(r^2 - 6r + 9 = 0\)
Complex roots
\(r^2 + 4 = 0\)

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Subsection Fundamental Solutions

To solve (34), we begin by forming the characteristic equation as we did in the previous section. For this case, the characteristic equation is the quadratic

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\begin{equation} a r^2 + b r + c = 0\tag{35} \end{equation}

Solving this equation using the quadratic formula yields:

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\begin{equation} r_1 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}, \quad r_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\tag{36} \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.

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Subsection The General Solution

Once we have the fundamental solutions, the superposition principle tells us that

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\begin{equation*} y = c_1 e^{r_1 x} + c_2 e^{r_2 x} \end{equation*}

is a solution to (34) for any constants \(c_1\) and \(c_2\text{.}\)

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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.

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This issue relates to linear dependence, which also depends on the interval where the solutions are defined.

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To create a second, independent solution, we multiply the second term by \(x\text{,}\) giving:

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\begin{equation*} y = c_1 e^{r x} + c_2 x e^{r x}. \end{equation*}

This guarantees complete general solution with independent terms. There are three distinct scenarios to consider, based on the discriminant of the characteristic equation:

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\begin{equation*} \Delta = b^2 - 4ac \end{equation*}

Each case leads to a different form of the solution, broken down as follows.

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Subsection Case 1 \((\Delta > 0)\text{:}\) Real, Non-Repeating Solutions

If \(\Delta\) is positive, then characteristic equation has the two unequal real solutions:

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\begin{equation*} r_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad r_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\text{,} \end{equation*}

so the general solution is

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\begin{equation*} y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\text{.} \end{equation*}

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Subsection Case 2 \((\Delta < 0)\text{:}\) Two Complex Solutions

If \(\Delta\) is negative, the characteristic equation has two complex conjugate solutions:

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\begin{equation*} r_1 = \alpha + i \beta, \qquad r_2 = \alpha - i \beta \end{equation*}

The corresponding general solution is:

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\begin{equation*} y = c_1 e^{(\alpha + i \beta)x} + c_2 e^{(\alpha - i \beta)x} \end{equation*}

However, it is common to express this using sine and cosine (via Euler’s Formula):

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\begin{equation*} y = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x)) \end{equation*}

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Subsection Case 3 \((\Delta = 0)\text{:}\) Real, Repeating Solutions

If \(\Delta\) is zero, then the characteristic equation has a repeated root:

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\begin{equation*} r_1 = r_2 = \frac{-b}{2a} \end{equation*}

In this case, the general solution is:

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\begin{equation*} y = c_1 e^{r x} + c_2 x e^{r x} \end{equation*}

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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.

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✳️ 2nd-Order LHCC General Solution Cases.

Let \(r_1\text{,}\) \(r_2\) be the roots of the characteristic equation for a \(2\)nd-order LHCC equation.

Case 1 \((\Delta > 0)\) \(\small \text{unequal real roots}\) 🔗: \(r_1 \ne r_2\)
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\(\ds y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)
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Case 2 \((\Delta < 0)\) \(\small \text{complex roots}\) 🔗: \(r = \alpha \pm i\beta\)
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\(\ds y = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\)
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Case 3 \((\Delta = 0)\) \(\small \text{repeated real root}\) 🔗: \(r_1 = r_2\)
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\(\ds y = c_1 e^{r x} + c_2 x e^{r x}\)
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Checkpoint 140. 📖❓ Match the Characteristic & General Solutions.

Match the characteristic solutions to the corresponding general solution.

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\(\small r = 5 \)
\(\small C e^{5x} \)
\(\small r_1 = 2, \ r_2 = 3 \)
\(\small C_1 e^{2x} + C_2 e^{3x} \)
\(\small r_1 = -1, \ r_2 = -4 \)
\(\small C_1 e^{-x} + C_2 e^{-4x} \)
\(\small r_1 = 3 + 2i, \ r_2 = 3 - 2i \)
\(\small e^{3x} (C_1 \cos(2x) + C_2 \sin(2x)) \)
\(\small r_1 = -2, \ r_2 = -2 \)
\(\small (C_1 + C_2 x) e^{-2x} \)
\(\small r_1 = -1 + i, \ r_2 = -1 - i \)
\(\small e^{-x} (C_1 \cos(x) + C_2 \sin(x)) \)

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🌌 Example 141. Case 1 Examples.

Find the general solution to each second-order LHCC equation.

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\(\ds y'' - 5y' + 6y = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - 5r + 6 = 0 \end{equation*}

Factoring gives:

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\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:

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\begin{equation*} y = c_1 e^{2x} + c_2 e^{3x} \end{equation*}

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\(\ds 5y'' + 13y' - 2y = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} 5r^2 + 13r - 2 = 0 \end{equation*}

Using the quadratic formula:

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\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:

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\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 \(r_2 = \frac{-13 + \sqrt{201}}{10}\text{.}\)

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🌌 Example 142. Case 2 Examples.

Find the general solution to each second-order LHCC equation.

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\(\ds \omega'' + 4\omega' + 5\omega = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r^2 + 4r + 5 = 0 \end{equation*}

Applying the quadratic formula:

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\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:

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\begin{equation*} \omega(t) = e^{-2t}(c_1 \cos t + c_2 \sin t) \end{equation*}

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\(\ds y'' + 25y = 0\)

Solution.

The characteristic equation is:

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\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:

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\begin{equation*} y = e^{0x}\left(c_1\cos(5x) + c_2\sin(5x)\right) \end{equation*}

or simply:

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\begin{equation*} y = c_1 \cos(5x) + c_2 \sin(5x) \end{equation*}

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🌌 Example 143. Case 3 Examples.

Find the general solution to each second-order LHCC equation.

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\(\ds y'' + 4y' + 4y = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r^2 + 4r + 4 = 0 \end{equation*}

Factoring gives:

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\begin{equation*} (r + 2)^2 = 0 \end{equation*}

So \(r = -2\) is a repeated root, and the general solution is:

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\begin{equation*} y = c_1 e^{-2x} + c_2 x e^{-2x} \end{equation*}

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\(\ds 16\frac{d^2q}{dt^2} - 8\frac{dq}{dt} + q = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} 16r^2 - 8r + 1 = 0 \end{equation*}

Using the quadratic formula:

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\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:

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\begin{equation*} q(t) = c_1 e^{\frac{1}{4}t} + c_2 t e^{\frac{1}{4}t} \end{equation*}

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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.

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

Second-order LHCC equations have the form \(a y'' + b y' + c y = 0\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are constants.

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The solution process begins by forming the characteristic equation \(a r^2 + b r + c = 0\text{.}\)

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The roots of the characteristic equation determines the form of the general solution as outlined in ✳️ 2nd-Order LHCC General Solution Cases.

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