Constructing a matrix that computes derivatives Consider the subset of functions given by $S = \text{Span}(e^{2t}\, \sin\, 3t, e^{2t}\, \cos\, 3t)$:
Show that the derivatives of $e^{2t}\, \sin\, 3t$ and $e^{2t}\, \cos\, 3t$ are also in $S$ and then use this fact to construct a matrix $D$ that computes derivatives of functions in $S$.
Use the inverse of $D$ to find the integrals of $e^{2t}\, \sin\, 3t$ and $e^{2t}\, \cos\, 3t$.
 A: Let
$$
v(t)=e^{2t}\sin 3t \qquad w(t)=e^{2t}\cos 3t
$$
and note that $S=\DeclareMathOperator{Span}{Span}\Span\{v,w\}$. That is, the functions in $S$ are exactly those functions of the form
$$
\lambda_1\cdot v(t)+\lambda_2\cdot w(t)\tag{1}
$$
Differentiating (1) with respect to $t$ gives
$$
(2\lambda_1-3\lambda_2)\cdot v(t)+(3\lambda_1+2\lambda_2)\cdot w(t)
$$
(Check this on your own!)
Thus our differentiation operator $T:S\to S$ is defined on the basis $\beta=\{v,w\}$ by
$$
\begin{array}{rcrcr}
T(v) &=& \color{red}{2}\cdot v  &+& \color{red}{3}\cdot w \\
T(w) &=& \color{blue}{-3}\cdot v &+& \color{blue}{2}\cdot w
\end{array}
$$
The matrix of this linear operator with respect to $\beta$ is
$$
D=
\begin{bmatrix}
\color{red}{2} & \color{blue}{-3} \\
\color{red}{3} & \color{blue}{2}
\end{bmatrix}
$$
Can you find the inverse of $D$ and finish the problem?
A: Step 1: Can we show that the derivatives of those functions are in the span?


*

*Is $$\frac{d}{dt} e^{2t}\sin 3t = \alpha e^{2t}\sin 3t + \beta e^{2t}\cos 3t$$ for some $\alpha, \beta$?

*Is $$\frac{d}{dt} e^{2t}\cos 3t = \alpha e^{2t}\sin 3t + \beta e^{2t}\cos 3t$$ for some $\alpha, \beta$?


Surely you can apply the product rule to show that it is in fact true.
Step 2: Given that$$\frac{d}{dt} e^{2t}\sin 3t = \alpha e^{2t}\sin 3t + \beta e^{2t}\cos 3t$$ for some $\alpha, \beta$, what does the matrix look like?
We have
$$D\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 3\end{pmatrix}$$
and
$$D\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ 2\end{pmatrix}.$$
Can you construct the matrix D?
