Section C.4 Linear Homogeneous Constant Coefficients

N-th Derivative of $e^{r x}$ .

e^{r x}

r,

\underset{rule}{chain} \to {[e^{r x}]}^{'} = r e^{r x}, {[e^{r x}]}^{″} = r^{2} e^{r x}, \dots, {[e^{r x}]}^{(n)} = r^{n} e^{r x} .

2nd Order LHCC Complex Case 3.

The following explains how Case 3 (complex) comes directly from Case 1 (real &

\neq

Since

r_{1}

and

r_{2}

are complex, they can be written as

r_{1} = α + β i and r_{2} = α - β i .

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{aligned} y & = C_{1} e^{(α + β i) x} + C_{2} e^{(α - β i) x} \\ = C_{1} e^{α x} e^{i β x} + C_{2} e^{α x} e^{- i β x} \\ = e^{α x} [C_{1} e^{i β x} + C_{2} e^{- i β x}] \\ = e^{α x} [C_{1} (\cos (β x) + i \sin (β x)) + C_{2} (\cos (- β x) + i \sin (- β x))] \\ = e^{α x} [C_{1} \cos (β x) + i C_{1} \sin (β x) + C_{2} \cos (- β x) + i C_{2} \sin (- β x)] \\ = e^{α x} [C_{1} \cos (β x) + i C_{1} \sin (β x) + C_{2} \cos (β x) - i C_{2} \sin (β x)] \\ = e^{α x} [(\underset{C_{1}}{\underset{⏟}{C_{1} + C_{2}}}) \cos (β x) + \underset{C_{2}}{\underset{⏟}{i (C_{1} - C_{2})}} \sin (β x)] \\ = e^{α x} (C_{1} \cos (β x) + C_{2} \sin (β x)) . \end{aligned}

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^{- 3 x}

terms below are pairs of like terms, which can be combined as follows:

\begin{matrix} \underset{―}{3 x^{2}} + \underset{―}{\underset{―}{5 e^{- 3 x}}} - 2 + \underset{―}{7 x^{2}} - \underset{―}{\underset{―}{4 e^{- 3 x}}} \\ \underset{―}{10 x^{2}} + \underset{―}{\underset{―}{e^{- 3 x}}} - 2 . \end{matrix}

Polynomial Factoring Calculator.

Figure C.10. Type in any polynomial of

r .

Use “

*

” for multiplication. If the polynomial has rational roots, it will show the factors.

To find the characteristic equation, we substitute

y = e^{r x}

into the

n

-th order LHCC equation and solve for

r .

Doing this gives us

\begin{aligned} a_{n} {[e^{r x}]}^{(n)} + \dots + a_{2} {[e^{r x}]}^{″} + a_{1} {[e^{r x}]}^{'} + a_{0} e^{r x} = & 0 \\ a_{n} r^{n} e^{r x} + \dots + a_{2} r^{2} e^{r x} + a_{1} r e^{r x} + a_{0} e^{r x} = & 0 \\ (a_{n} r^{n} + \dots + a_{2} r^{2} + a_{1} r + a_{0}) e^{r x} = & 0 . \end{aligned}

Since

e^{r x}

is never zero, we must have

a_{n} r^{n} + \dots + a_{2} r^{2} + a_{1} r + a_{0} = 0

which is an

n

-th order polynomial in

r .

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Section C.4 Linear Homogeneous Constant Coefficients

N-th Derivative of erx.

2nd Order LHCC Complex Case 3.

Like Terms.

Polynomial Factoring Calculator.

N-th Derivative of $e^{r x}$ .