Section 6.2 Solutions
Before diving into how to solve
LHCC equations, let’s develop some intuition about what the solutions should look like. For this discussion we will assume that the equations contain a
\(y\) term. If they didn’t, then you could repeatedly integrate both sides of the equation until you are left with a
\(y\) term and a nonhomogeneous equation.
For example, the following shows how an LHCC equation with no
\(y\) term can be reduced to a nonhomogeneous equation with a
\(y\) term.
\begin{align*}
y''' - 2y'' = 0 \ \ \rightarrow \ \ (y' - 2y)'' =\amp\ 0 \quad \text{(direct integration x2)} \\
(y' - 2y)' =\amp\ c_1 \\
y' - 2y =\amp\ c_1x + c_2 \text{.}
\end{align*}
We begin with the simple first-order LHCC equation,
\begin{equation}
y' - 2 y = 0\tag{26}
\end{equation}
and note that this equation requires
\(y' - 2y\) to sum to zero. Recall from algebra that only
like terms can combine to simplify to zero. So, for the equation to hold,
\(y'\) and
\(y\) must be like terms.
Let’s test a few guesses for \(y\text{:}\) \(y = x^2\text{,}\) \(y = 2 \ln x\text{,}\) and \(y = e^{2x}\text{.}\)
\begin{gather*}
\left[x^2\right]' - 2 \left[x^2\right] = 0 \\
\ub{2x - 2x^2}_{\text{not like terms}} = 0
\end{gather*}
\begin{gather*}
\left[2\ln x\right]' - 4\ln x = 0 \\
\ub{\frac{2}{x} - 4\ln x}_{\text{not like terms}} = 0
\end{gather*}
\begin{gather*}
\left[e^{2x}\right]' - 2 \left[e^{2x}\right] = 0 \\
\ub{2e^{2x} - 2e^{2x}}_{\text{like terms}\, =\, 0} = 0
\end{gather*}
As you can see, only \(y = e^{2x}\) produces like terms that simplify to zero. This is why exponential functions often work well in LHCC equations. Let’s now examine a more complex example to expand on this idea.
Example 4.
\(\ \ \)
\begin{equation*}
y'' - 3y' - 4y = 0.
\end{equation*}
Verify that \(y = e^{4x}\) is a solution, but \(y = x^4\) is not.
Solution.
Let’s check each function:
\begin{align*}
[e^{4x}]'' - 3\left[e^{4x}\right]' - 4\left[e^{4x}\right] =\amp\ 0 \\
\us{\text{like terms simplify to }0}{\ub{
16e^{4x} - 12e^{4x} - 4e^{4x}
}}
=\amp\ 0 \\
0 =\amp\ 0 \ \ ✅
\end{align*}
\begin{align*}
[x^4]'' - 3\left[x^4\right]' - 4\left[x^4\right] =\amp\ 0 \\
\us{\text{not like terms, do not combine}}{\ub{
12x^2 - 12x^3 - 4x^4
}}
=\amp\ 0 \ \ ❌
\end{align*}
Therefore, only \(y = e^{-4x}\) is a solution.
This example further supports the claim that exponential functions, like \(e^{rx}\text{,}\) are well-suited for LHCC equations because they produce like terms when differentiated. Later, we will see that if \(e^{rx}\) is a solution then, in some cases, it is possible for \(xe^{rx}\) to be a solution as well.
We conclude this section with an important concept that will guide us in the sections to follow.
Terms of the Solution to an LHCC Equation.
The general solution to an LHCC equation involves terms of the form:
\begin{equation*}
e^{rx}, \quad \ub{xe^{rx}, \quad x^2e^{rx}, \quad x^3e^{rx},\ \ldots}_{\text{For Special Cases}}
\end{equation*}
where
\(r\) is a constant. Note, if
\(r\) is complex, then
\(r=a+\theta i\text{,}\) and by
Euler’s formula,
\begin{equation*}
e^{rx} = e^{(a+\theta i)x} = e^{ax}\cdot e^{\theta i x} = e^{ax}(\cos(bx) + i\sin(bx)) \text{.}
\end{equation*}
In this case, it is common to see a solution written with sine and cosine terms.
Reading Questions Check-Point Questions
1. \(e^{rx} \) is the only function whose derivative is a multiple of itself.
\(e^{rx} \) is the only function whose derivative is a multiple of itself- True
- Correct!
- False
- Incorrect.
2. \(e^{rx} \) and its derivatives are "like terms" that can be combined.
\(e^{rx} \) and its derivatives are "like terms" that can be combined- True
- Correct!
- False
- Incorrect.
3. The hundredth derivative of \(e^{7x} \) is equal to a constant times \(e^{7x} \).
The hundredth derivative of \(e^{7x} \) is equal to a constant times \(e^{7x} \)- True
- Correct!
- False
- Incorrect.
4. Without solving it, which function satisfies the property \(y' = -2y \text{?}\)
Without solving it, which function satisfies the property \(y' = -2y \text{?}\)- \(y = e^{2x} \)
- Incorrect.
- \(y = e^{-2x} \)
- Correct! The derivative of \(y = e^{-2x} \) is \(-2\) times itself.
- \(y = \sin(-2x) \)
- Incorrect.
- \(y = -x^2 \)
- Incorrect.
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