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).

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

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?

  • $\begingroup$ I edited the formatting in your post. Please check it still expresses what you intended to say. $\endgroup$ – Christoph Oct 28 '13 at 9:52
  • $\begingroup$ How do you see about your current solution? It seems correct to me. $\endgroup$ – Shuchang Oct 28 '13 at 10:20
  • $\begingroup$ 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. $\endgroup$ – user507974 Oct 28 '13 at 10:23
  • $\begingroup$ Notice how $A$ is sent to the second row with a factor of 2. And similarly for the other elements of the vector. $\endgroup$ – Brady Trainor Oct 28 '13 at 10:25
  • $\begingroup$ On the other end of the problem, I think we want the sum of the $n=2$ and $n=0$ case. $\endgroup$ – 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, $$\begin{pmatrix} A\\0\\0\\0 \end{pmatrix}\mapsto \begin{pmatrix} 0\\2A\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\B\\0\\0 \end{pmatrix}\mapsto \begin{pmatrix} -2B\\0\\0\\0 \end{pmatrix},$$

etc. In a sense, it may be easier in this situation to consider what happens to the basis vectors. This gives the columns of your transformation matrix.

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