# Undetermined Coefficients System

#### Metronome

##### New member
I want to solve the system

$$\displaystyle x' = 4x - 13y + 2e^{2t}\cos(3t) \\ y' = x$$
using undetermined coefficients. For the associated homogeneous solution, I get $$\displaystyle \vec{x_h} = Ae^{2t}\begin{bmatrix}13\cos(3t) \\ 2\cos(3t) + 3\sin(3t)\end{bmatrix} + Be^{2t}\begin{bmatrix}13\sin(3t) \\ 2\sin(3t) - 3\cos(3t)\end{bmatrix}$$.

My first guess for the particular solution was $$\displaystyle \vec{x_p} = \begin{bmatrix}a_1 \\ a_2\end{bmatrix}e^{2t}\cos(3t) + \begin{bmatrix}b_1 \\ b_2\end{bmatrix}e^{2t}\sin(3t)$$.

This was a risky guess, since it closely resembles parts of $$\displaystyle \vec{x_h}$$, but I don't know the exact conditions under which the default guess fails when working with systems. Sure enough, I ended up with an inconsistent system for $$\displaystyle a_1$$, $$\displaystyle a_2$$, $$\displaystyle b_1$$, and $$\displaystyle b_2$$ as expected.

I looked at the math I used to solve the corresponding second-order equation, $$\displaystyle y'' - 4y' + 13y = 2e^{2t}\cos(3t)$$, and saw that the guess $$\displaystyle y_p = ate^{2t}\cos(3t) + bte^{2t}\sin(3t)$$ was successful, so my next guess was $$\displaystyle \vec{x_p} = \begin{bmatrix}a_1 \\ a_2\end{bmatrix}te^{2t}\cos(3t) + \begin{bmatrix}b_1 \\ b_2\end{bmatrix}te^{2t}\sin(3t)$$.

I used the exponential shift rule (which I'm assuming/hoping is valid for systems?) to get $$\displaystyle e^{2t}(D + 2)\left[\begin{bmatrix}a_1 \\ a_2\end{bmatrix}t\cos(3t) + \begin{bmatrix}b_1 \\ b_2\end{bmatrix}t\sin(3t)\right] = \begin{bmatrix}4 & -13 \\ 1 & 0\end{bmatrix}\left(\begin{bmatrix}a_1 \\ a_2\end{bmatrix}te^{2t}\cos(3t) + \begin{bmatrix}b_1 \\ b_2\end{bmatrix}te^{2t}\sin(3t)\right) + \begin{bmatrix}2e^{2t}\cos(3t) \\ 0\end{bmatrix},$$ but this once again led to an inconsistent algebraic system, namely, $$\displaystyle \begin{cases}3b_1 + 2a_1 = 4a_1 - 13a_2 \\ -3a_1 + 2b_1 = 4b_1 - 13b_2 \\ a_1 = 2 \\ b_1 = 0 \\ 3b_2 +2a_2 = a_1 \\ -3a_2 + 2b_2 = b_1 \\ a_2 = 0 \\ b_2 = 0\end{cases}$$.

After computing the solution with Wolfram, it appears that the required guess was the sum of the two guesses above, $$\displaystyle \vec{x_p} = \begin{bmatrix}a_1 \\ a_2\end{bmatrix}te^{2t}\cos(3t) + \begin{bmatrix}b_1 \\ b_2\end{bmatrix}te^{2t}\sin(3t) + \begin{bmatrix}c_1 \\ c_2\end{bmatrix}e^{2t}\cos(3t) + \begin{bmatrix}d_1 \\ d_2\end{bmatrix}e^{2t}\sin(3t)$$, but this form seems to break the general pattern. For example, if the forcing term in a second-order linear ODE is $$\displaystyle e^t$$, and $$\displaystyle y_h = e^t$$ is an associated homogeneous solution, then the guess $$\displaystyle y_p = ate^t$$ is recommended; you don't need $$\displaystyle y_p = ate^t + be^t$$. The corresponding second-order equation above serves as another example. According to this answer, we can conclude that the guess for systems where all but one component of the forcing term are $$\displaystyle 0$$ should be the same as if that non-zero component were the forcing term of a single linear ODE.

Why does this system, by contrast, require the original terms to be kept when the new terms with an extra factor of $t$ are introduced? Additionally, why does the corresponding second-order equation, $$\displaystyle y'' - 4y' + 13y = 2e^{2t}\cos(3t)$$, provide misleading information about what the guess should be?

#### HallsofIvy

##### Elite Member
Here is a slightly different way of looking at it:
Your first equation is $$\displaystyle x′=4x−13y+2e^{2t}cos(3t)$$. Differentiating that again, $$\displaystyle x''= 4x'- 13y'+ 4e^{2t}cos(3t)- 6e^{2t}sin(3t)$$. Your second equation is $$\displaystyle y'= x$$ so that can be written $$\displaystyle x''= 4x'- 13x+ 4e^{2t}cos(3t)- 6e^{2t}sin(3t)$$, a second degree equation for x(t) only.

That has associated homogeneous equation $$\displaystyle x''- 4x'+ 13x= 0$$ which has characteristic equation $$\displaystyle r^2- 4r+ 13= 0$$ and characteristic roots $$\displaystyle r= 2\pm 3i$$. So the associated homogenous equation has general solution $$\displaystyle x(t)= e^{2x}(Acos(3x)+ Bsin(3x))$$.

Now, the point is that the "non-homogeneous part", $$\displaystyle 2e^{2t}cos(3t)$$ has those same functions. Since they will make the equation 0, there is NO choice of constants that will make them equal to the "non-homogeneous" part of the equation. I would think that well before you had reached this point in you class you had learned that you "needed to multiply buy t" to fix that.

#### Metronome

##### New member
Yeah, I was able to solve the equation with more or less this approach. I just wasn't sure how to see what the guess should be in advance when it's in system form.