Section C.4 Linear Homogeneous Constant Coefficients
N-th Derivative of \(e^{rx}\).
\(e^{rx}\)\(r\text{,}\)
\begin{equation*}
\us{\ \text{ rule}}{\text{chain}} \rightarrow \ \ \left[e^{rx}\right]' = r e^{rx}, \quad \left[e^{rx}\right]'' = r^2 e^{rx}, \quad \ldots, \quad \left[e^{rx}\right]^{(n)} = r^n e^{rx}\text{.}
\end{equation*}
2nd Order LHCC Complex Case 3.
The following explains how Case 3 (complex) comes directly from Case 1 (real & \(\ne\)).
Since \(r_1\) and \(r_2\) are complex, they can be written as
\begin{equation*}
r_1 = \alpha + \beta i \quad \text{and} \quad r_2 = \alpha - \beta i\text{.}
\end{equation*}
Substituting these into (30), using Euler’s Formula, the even property of cosine and odd property of sine, we can rewrite the general solution as
\begin{align*}
y\ &= C_1 e^{(\alpha + \beta i) x} + C_2 e^{(\alpha - \beta i) x} \\
&= C_1 e^{\alpha x} e^{i \beta x} + C_2 e^{\alpha x} e^{-i \beta x} \\
&= e^{\alpha x} \Big{[}C_1 e^{i \beta x} + C_2 e^{-i \beta x}\Big{]} \\
&= e^{\alpha x} \Big{[}C_1 (\cos(\beta x) + i\sin(\beta x)) + C_2 (\cos(-\beta x) + i\sin(-\beta x))\Big{]} \\
&= e^{\alpha x} \Big{[}C_1 \cos(\beta x) + i C_1 \sin(\beta x) + C_2 \cos(-\beta x) + i C_2 \sin(-\beta x)\Big{]} \\
&= e^{\alpha x} \Big{[}C_1 \cos(\beta x) + i C_1 \sin(\beta x) + C_2 \cos(\beta x) - i C_2 \sin(\beta x)\Big{]} \\
&= e^{\alpha x} \Big{[}(\us{C_1}{\ub{C_1 + C_2}}) \cos(\beta x) + \us{C_2}{\ub{i (C_1 - C_2)}} \sin(\beta x)\Big{]} \\
&= e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x)) \text{.}
\end{align*}
Like Terms.
Terms are called like terms if they have identical variable parts. That is, they differ only by a coefficient. Like terms can be combined via addition and subtraction. For example, the \(x^2\) and \(e^{-3x}\) terms below are pairs of like terms, which can be combined as follows:
\begin{gather*}
\underline{3x^2} + \underline{\underline{5e^{-3x}}} - 2 + \underline{7x^2} - \underline{\underline{4e^{-3x}}}\\
\underline{10x^2} + \underline{\underline{e^{-3x}}} - 2 \text{.}
\end{gather*}
Polynomial Factoring Calculator.
To find the characteristic equation, we substitute \(y = e^{rx}\) into the \(n\)-th order LHCC equation and solve for \(r\text{.}\) Doing this gives us
\begin{align*}
a_n\ \left[e^{rx}\right]^{(n)} + \cdots + a_2\ \left[e^{rx}\right]'' + a_1\ \left[e^{rx}\right]' + a_0\ e^{rx} =\amp\ 0 \\
a_n\ r^n e^{rx} + \cdots + a_2\ r^2 e^{rx} + a_1\ r e^{rx} + a_0\ e^{rx} =\amp\ 0 \\
\left(a_n\ r^n + \cdots + a_2\ r^2 + a_1\ r + a_0\right)e^{rx} =\amp\ 0 \text{.}
\end{align*}
Since \(e^{rx}\) is never zero, we must have
\begin{equation*}
a_n\ r^n + \cdots + a_2\ r^2 + a_1\ r + a_0 = 0
\end{equation*}
which is an \(n\)-th order polynomial in \(r\text{.}\)
You have attempted of activities on this page.
