# Understanding how to take derivatives with matrices

Currently we are doing 2nd order differential equations (we already did systems of homogenous two first order equations) and now that we have non-homogenous 2nd order equations we are doing method of undetermined coefficients. I really liked some of the linear algebra-based introduction to the material the instructor gave, and kind of wanted to apply more linear algebra to this problem. I was thinking I wanted to create a matrix to represent taking the derivative of the nonhomogenous equation as I set up the particular solution, just to kind of help me relate both together, but for some reason every time i try and set up the matrices for it I feel like I am getting it wrong or misunderstanding something if i do get it right. For example $y''+y= 12\sin 2t + 4t\cos 2t$ I know there is there is "multiplicity" on the right side so i need $A\sin 2t + B\cos 2t + Ct\sin 2t + Dt\sin 2t$ for my coefficients. I get the feeling I'm doing something wrong with setting up this matrix. The rows correspond to $\sin 2t$, $\cos 2t$, $t\sin 2t$, $t\cos 2t$ (i think).

$$\begin{pmatrix} 0 & -2 & 1 & 0 \\ 2 & 0 & 0 & 1 \\ 0 & 0 & 0 & -2 \\ 0 & 0 & 2 & 0 \\ \end{pmatrix}^n \begin{pmatrix} A \\ B \\ C \\ D \end{pmatrix} = \text{nth derivative of } \begin{pmatrix} A \\ B \\ C \\ D \end{pmatrix}$$

the answer for $n=1$ was $[-2B+C,2A+D,2C,-2D]$, which i believe is right. This was not my first attempt and I did a lot of trial and error to get here, which bugs me. I wanted some help interpreting what exactly was going on, and what columns/rows in this operation really were corresponding to on the calculus end. Am I just approaching this problem incorrectly and misunderstanding how to take the derivative with a matrix?

• I edited the formatting in your post. Please check it still expresses what you intended to say. – Christoph Oct 28 '13 at 9:52
• How do you see about your current solution? It seems correct to me. – Shuchang Oct 28 '13 at 10:20
• I tried several different ones until it actually outputted the correct answer that i got by traditional methods, but i still dont really understand how this mechanism is actually working. – user507974 Oct 28 '13 at 10:23
• Notice how $A$ is sent to the second row with a factor of 2. And similarly for the other elements of the vector. – Brady Trainor Oct 28 '13 at 10:25
• On the other end of the problem, I think we want the sum of the $n=2$ and $n=0$ case. – Brady Trainor Oct 28 '13 at 10:26

Consider that the derivative takes $\sin 2t$ to $2\cos t$. This corresponds with $$\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}\mapsto \begin{pmatrix} 0\\2\\0\\0 \end{pmatrix}.$$
This effectively tells you how the first column of your matrix must affect your coefficient $A$ in the first row of your vector. In other words, $$\begin{pmatrix} A\\0\\0\\0 \end{pmatrix}\mapsto \begin{pmatrix} 0\\2A\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\B\\0\\0 \end{pmatrix}\mapsto \begin{pmatrix} -2B\\0\\0\\0 \end{pmatrix},$$