DE-BOOK Exponential Solutions

Section Exponential Solutions

Now that we can identify LHCC equations, it’s time to ask a deeper question: what kinds of functions satisfy these equations? As it turns out, exponential functions are not just one possible answer, they are the natural solution.

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In this section, we’ll explore what makes exponentials so special. You’ll see why other functions fail to work, why exponentials succeed, and how this leads us to a powerful tool called the characteristic equation.

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Checkpoint 133. Derivatives & Like-Terms.

(a) ↩️☝ Recall: Functions that are their own derivatives.

What is the only function that is equal to its own derivative?

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\(\quad y = e^{x} \)
Correct!\(\frac{d}{dx}[e^{x}] = e^{x}\)
\(\quad y = \tan x \)
Incorrect. \(\frac{d}{dx}[\tan x] = \sec^2 x\text{,}\) not \(\tan x\text{.}\)
\(\quad y = x^2 \)
Incorrect. \(\frac{d}{dx}[x^2] = 2x\text{,}\) not \(x^2\text{.}\)
\(\quad y = \ln x \)
Incorrect. \(\frac{d}{dx}[\ln x] = \frac{1}{x}\text{,}\) not \(\ln x\text{.}\)

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(b) ↩️☝ Recall: Like Terms.

Terms are like-terms if they differ only by a coefficient and can be combined via addition and subtraction.

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For example, the pairs \(\{5e^{-3x}, 4e^{-3x}\}\) and \(\{3x^2, 7x^2\}\) are like-terms and can be combined/simplifed as shown here:

\begin{equation*} \underline{3x^2} + 2e^{7x} + \underline{\underline{5e^{-3x}}} - 2 + \underline{7x^2} - \underline{\underline{4e^{-3x}}} = \underline{10x^2} + 2e^{7x} + \underline{\underline{e^{-3x}}} - 2 \text{.} \end{equation*}

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Which of the following are like-terms with \(\cos(3x)\text{?}\)

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\(\quad 2\cos(3x) \)
Correct! \(2\cos(3x)\) is a like-term with \(\cos(3x)\) and \(2\cos(3x) + \cos(3x) = 3\cos(3x)\text{.}\)
\(\quad -7\cos(3x) \)
Correct! \(-7\cos(3x)\) is a like-term with \(\cos(3x)\) and \(-7\cos(3x) + \cos(3x) = -6\cos(3x)\text{.}\)
\(\quad 3\sin(3x) \)
Incorrect. \(3\sin(3x)\) is not a like-term with \(\cos(3x)\text{.}\)
\(\quad 4\cos(2x) \)
Incorrect. \(4\cos(2x)\) is not a like-term with \(\cos(3x)\text{.}\)
\(\quad 3x \)
Incorrect. \(3x\) is not a like-term with \(\cos(3x)\text{.}\)

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(c) ↩️☝ Recall: Functions that are like terms with their own derivatives.

Which of the functions is a like-term with its own derivative?

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\(\quad y = e^{10x} \)
Correct!\(\frac{d}{dx}[e^{10x}] = 10e^{x}\ \Rightarrow\ e^{10x}\) & \(10e^{10x}\) ARE like terms and can be combined by addition.
\(\quad y = \sin (4x) \)
Incorrect. \(\frac{d}{dx}[\sin (4x)] = 4\cos (4x)\ \Rightarrow\ \sin(4x)\) & \(4\cos (4x)\) ARE NOT like terms and cannot be combined by addition.
\(\quad y = x^4 \)
Incorrect. \(\frac{d}{dx}[x^4] = 4x^3\ \Rightarrow\ x^4\) & \(4x^3\) ARE NOT like terms and cannot be combined by addition.
\(\quad y = x^{-1} \)
Incorrect. \(\frac{d}{dx}[x^{-1}] = -x^{-2}\ \Rightarrow\ x^{-1}\) & \(-x^{-2}\) ARE NOT like terms and cannot be combined by addition.

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Subsection Motivating the Exponential Solution

Consider the simple linear homogeneous constant coefficient differential equation:

\begin{equation*} y' - 2y = 0\text{.} \end{equation*}

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For the moment, forget that it is a differential equation and imagine that it is just two functions that simplify to zero after subtraction. That is, they cancelled out with each other. Algebra tells you that only like-terms simplify via subtraction.

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For example, \(5x - 3x - 2x = 0\) since all three are like-terms. However, \(5x - 3x^2 - 2y\) does not simplify at all since none are like-terms.

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Logically, if \(y\) is the solution, then this equation says \(y\) and \(y'\) must be like-terms since they subtract to zero. This leads to the next big question:

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“What function is a like-term with its derivative?”

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You may recall from calculus that the derivative of an exponential function is another exponential function. For example, if \(y=e^{3x}\text{,}\) then \(y'=3e^{3x}\text{,}\) which are like-terms. This unique property makes exponentials ideal candidates for solutions.

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To illustrate this, let’s substitute different functions into a first-order LHCC equation and see which ones cancel.

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Case 1: \(\quad y = x^2\)

\begin{align*} \left[x^2\right]' - 2 \left[x^2\right] \amp = 0 \\ \ub{2x - 2x^2}_{\text{non like-terms}} \amp \ne 0 \end{align*}

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Case 2: \(\quad y = \ln x\)

\begin{align*} \left[\ln x\right]' - 2\left[\ln x\right] \amp = 0 \\ \ub{\frac{1}{x} - 2\ln x}_{\text{non like-terms}} \amp \ne 0 \end{align*}

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Case 3: \(\quad y = e^{2x}\)

\begin{align*} \left[e^{2x}\right]' - 2 \left[e^{2x}\right] \amp = 0 \\ \ub{2e^{2x} - 2e^{2x}}_{\text{like-terms}} \amp = 0 \end{align*}

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Only the exponential function produces like-terms that cancel perfectly. Again, suggesting that exponential functions are ideal solutions to LHCC equations.

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Checkpoint 134. 📖❓ Valid Function Choices.

Consider the differential equation,

\begin{equation*} y' + by = 0 \end{equation*}

where \(b\) is a constant.

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Logically explain why it is impossible for \(x^3\) to be a solution to this equation no matter what value \(b\) takes.

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🌌 Example 135. Multiple Exponential Solutions.

Verify that \(y = e^{4x}\) and \(y = e^{-x}\) are solutions to the differential equation

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\begin{equation*} y'' - 3y' - 4y = 0 \end{equation*}

but that \(y = x^4\text{,}\) \(y = \ln x\text{,}\) and \(y = e^{3x}\) are not.

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Solution.

First, let’s check the first two exponential functions:

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\begin{equation*} y=e^{4x}: \end{equation*}

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\begin{align*} [e^{4x}]'' - 3\left[e^{4x}\right]' - 4\left[e^{4x}\right] \amp = 0 \\ \us{\large\text{like-terms simplify to }0}{\ub{ 16e^{4x} - 12e^{4x} - 4e^{4x} }} \amp = 0 \\ 0 \amp = 0 \ \ ✅ \end{align*}

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\begin{equation*} y=e^{-x}: \end{equation*}

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\begin{align*} [e^{-x}]'' - 3\left[e^{-x}\right]' - 4\left[e^{-x}\right] \amp = 0 \\ \us{\large\text{like-terms simplify to }0}{\ub{ e^{-x} + 3e^{-x} - 4e^{-x} }} \amp = 0 \\ 0 \amp = 0 \ \ ✅ \end{align*}

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As you can see, the exponential functions produce like-terms that cancel out and simplify to zero.

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Let’s see what happens when we plug in non-exponential functions:

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\begin{equation*} y=x^4: \end{equation*}

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\begin{align*} [x^4]'' - 3\left[x^4\right]' - 4\left[x^4\right] \amp = 0 \\ \us{\large\text{not like-terms, do not combine}}{\ub{ 12x^2 - 12x^3 - 4x^4 }} \amp = 0 \ \ ❌ \end{align*}

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\begin{equation*} y=\ln x: \end{equation*}

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\begin{align*} [\ln x]'' - 3\left[\ln x\right]' - 4\left[\ln x\right] \amp = 0 \\ \us{\large\text{not like-terms, do not combine}}{\ub{ -x^{-2} - 3x^{-1} - 4\ln x }} \amp = 0 \ \ ❌ \end{align*}

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In this case, the non-exponential functions produce non-like-terms that cannot be simplified and are not solutions.

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Finally, let’s check the exponential function, \(y = e^{3x}\text{:}\)

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\begin{equation*} y=e^{3x}: \end{equation*}

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\begin{align*} [e^{3x}]'' - 3\left[e^{3x}\right]' - 4\left[e^{3x}\right] \amp = 0 \\ \us{\large\text{like-terms don't simplify to }0}{\ub{ 9e^{3x} - 9e^{3x} - 4e^{3x} }} \amp = 0 \\ - 4e^{3x} \amp = 0 \ \ ❌ \end{align*}

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In this case, we get like-terms, but the coefficients do not sum to zero and the terms do not cancel. Therefore, \(y=e^{3x}\) is not a solution.

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These examples show that exponentials are the only functions that reliably cancel out to yield zero when substituted into an LHCC equation. Even among exponentials, only those with specific exponents will satisfy a given equation. Next, we’ll introduce the characteristic equation, an algebraic tool that helps us identify exactly which exponential functions will work.

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Subsection The Characteristic Equation

We now know that exponential functions \(y = e^{rx}\) are strong candidates for solutions to LHCC equations. But how do we find the values of \(r\) that work? The answer lies in forming a polynomial equation called the characteristic equation.

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Let’s try this with a few concrete examples. First, suppose

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\begin{equation*} y' - 5y = 0\text{.} \end{equation*}

Substituting \(y = e^{rx}\) gives

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\begin{align*} \left[e^{rx}\right]' - 5 e^{rx} \amp = 0 \\ r e^{rx} - 5 e^{rx} \amp = 0 \\ (r - 5)e^{rx} \amp = 0 \end{align*}

Since \(e^{rx} \ne 0\text{,}\) the equation reduces to

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\begin{equation*} r - 5 = 0 \quad \Rightarrow \quad r = 5 \end{equation*}

This algebraic equation, \(r - 5 = 0\text{,}\) is the characteristic equation. It tells us that \(y = e^{5x}\) is a solution.

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For second-order equations, we follow the same substitution process. For example, for

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\begin{equation*} y'' - 3y' - 4y = 0. \end{equation*}

we substitute \(y = e^{rx}\) into \(y\text{,}\) \(y'\text{,}\) and \(y''\text{,}\) resulting in

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\begin{align*} (r^2 e^{rx}) - 3 (r e^{rx}) - 4 (e^{rx}) \amp = 0 \\ (r^2 - 3r - 4) e^{rx} \amp = 0 \end{align*}

Setting the polynomial equal to zero gives the characteristic equation:

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\begin{equation*} r^2 - 3r - 4 = 0. \end{equation*}

Solving this gives

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\begin{equation*} (r - 4)(r + 1) = 0 \quad \Rightarrow \quad r = 4, \quad r = -1 \end{equation*}

so the general solution is

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\begin{equation*} y_1 = e^{4x}, \quad y_2 = e^{-x}. \end{equation*}

Checkpoint 136. 📖❓ Match the Characteristic Equation.

Match each characteristic equation to its differential equation.

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\(2r^2 + 5r - 1 = 0 \)
\(2y'' + 5y' - y = 0 \)
\(4r + 3 = 0 \)
\(4y' + 3y = 0 \)
\(r^2 + 1 = 0 \)
\(y'' + y = 0 \)
\(r^2 - r - 3 = 0\)
\(y'' - 3y = y'\)
\(r^2 - r = 0\)
\(y'' - y' = 0\)
\(y'' + y' - 3y = 0\)

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🌌 Example 137. Characteristic Equation & Exponential Solutions.

For each of the following equations, determine the characteristic equation and the exponential solutions that satisfy it.

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1. \(\ds\ 5y' + 6y = 0\)
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Solution.

Assuming \(y = e^{rx}\text{,}\) we have

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\begin{align*} 5\left[e^{rx}\right]' + 6\left[e^{rx}\right] = 0 \amp \\ 5r e^{rx} + 6 e^{rx} = 0 \amp \\ (5r + 6) e^{rx} = 0 \amp \text{.} \end{align*}

Thus, the characteristic equation is

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\begin{equation*} 5r + 6 = 0 \quad \Rightarrow \quad r = -\frac{6}{5}\text{,} \end{equation*}

and the only exponential solution is

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\begin{equation*} y = e^{-\frac{6}{5}x}. \end{equation*}

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2. \(\ds\ y'' - 4y = 0\)
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Solution.

Assuming \(y = e^{rx}\text{,}\) we have

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\begin{align*} r^2 e^{rx} - 4 e^{rx} = 0 \amp \\ (r^2 - 4) e^{rx} = 0 \amp \end{align*}

Thus, the characteristic equation is

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\begin{align*} r^2 - \amp 4 = 0 \\ \amp \Rightarrow\ (r - 2)(r + 2) = 0 \end{align*}

and the exponential solutions are

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\begin{equation*} y_1 = e^{2x}, \quad y_2 = e^{-2x}. \end{equation*}

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Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The derivative properties of exponential functions make them natural solutions for LHCC equations.

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Substituting \(y = e^{rx}\) into an LHCC equation leads to a polynomial equation in \(r\) called the characteristic equation.

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For each solution, \(r\text{,}\) that comes from the characteristic equation, you get a fundamental solution, \(e^{rx}\text{,}\) that satisfies the LHCC equation.

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Check Your Understanding.

Checkpoint 138. 🤔💭 Separation of Variables Reading Questions.

(a) 🤔💭 Exponential Like Terms.

The functions

\begin{equation*} e^{8x},\quad (e^{8x})',\quad (e^{8x})'',\quad (e^{8x})''' \end{equation*}

are all like-terms, which if added together, would simplify to one term.

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True
Correct! Computing the derivatives, you get
\begin{equation*} e^{8x},\quad 8e^{8x},\quad 64e^{8x},\quad 512e^{8x}\text{,} \end{equation*}
which are all like-terms.
False
Incorrect.

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