Differentiation of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x wrt x as Invertible Matrix

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Theorem

Let $a, b, x \in \R$ be real numbers.

Suppose $a \ne 0 \ne b$.

Denote $\ds f_1 = \map \exp {a x} \map \cos {b x}$, $f_2 = \map \exp {a x} \map \sin {b x}$.

Let $\map \CC \R$ be the space of continuous real-valued functions.

Let $\struct {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be the vector space of continuously differentiable real-valued functions.

Let $S = \span \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.

Let $D : S \to S$ be the derivative with respect to $x$.


Then, with respect to basis $\tuple {f_1, f_2}$, $D$ is expressible as:

$\mathbf D = \begin{bmatrix}

a & b \\ -b & a \end{bmatrix}$

and is invertible.


Proof

\(\ds \map D {f_1}\) \(=\) \(\ds \dfrac \d {\d x} \paren {\map \exp {a x} \map \cos {b x} }\)
\(\ds \) \(=\) \(\ds a \map \exp {a x} \map \cos {b x} - b \map \exp {a x} \map \sin {b x}\)
\(\ds \) \(=\) \(\ds a f_1 - b f_2\)
\(\ds \map D {f_2}\) \(=\) \(\ds \dfrac \d {\d x} \paren {\map \exp {a x} \map \sin {b x} }\)
\(\ds \) \(=\) \(\ds a \map \exp {a x} \map \sin {b x} + b \map \exp {a x} \map \cos{b x}\)
\(\ds \) \(=\) \(\ds b f_1 + a f_2\)

Thus:

\(\ds \mathbf D \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix}\) \(:=\) \(\ds \map D {\alpha_1 f_1 + \alpha_2 f_2}\)
\(\ds \) \(=\) \(\ds \alpha_1 \paren {a f_1 - b f_2} + \alpha_2 \paren {b f_1 + a f_2}\)
\(\ds \) \(=\) \(\ds \paren {a \alpha_1 + b \alpha_2} f_1 + \paren {-b \alpha_1 + a \alpha_2} f_2\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} a \alpha_1 + b \alpha_2 \\ -b \alpha_1 + a \alpha_2 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix}

a & b \\ -b & a \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix}\)

Furthermore:

$\map \det {\mathbf D} = a^2 + b^2$.

By Matrix is Invertible iff Determinant has Multiplicative Inverse, $\mathbf D$ is invertible.

$\blacksquare$


Sources

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