The Eigenvalue Method

Homogeneous

In all of our problems before, we have done problems with one differential equation. What if we have a series of differential equations? Here is an example. $$\begin{matrix}x'_1 & = & 3x_1 & & & + & x_3, \\ x'_2 & = & 9x_1 & - & x_2 & + & 2x_3, \\ x'_3 & = & -9x_1 & + & 4x_2 & - & x_3\end{matrix}$$ If you need a refresher on linear algebra and matrices in general go to mathwizurd.com/linalg. Anyway, we can rewrite that like this: $$\begin{bmatrix}x'_1 \\ x'_2 \\ x'_3\end{bmatrix} = \begin{bmatrix}3 & 0 & 1\\9 & - 1 & 2\\-9 & 4 & -1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$ Because the original problem is composed of constant coefficients, like what we did 3 posts before, we guess $e^{cx}$. So, for the entire system, we guess $\vec{v}e^{cx}$ If we plug it in into the general equation $x' = Ax$, we get $c\vec{v}e^{cx} = A\vec{v}e^{cx}$. We divide by $e^{cx}$ and we get $c\vec{v} = A\vec{v}$. We've seen this before: that's the definition of an eigenvalue eigenvector pair.

Once more, let's review how we got here. We started with $\vec{v}e^{cx}$. So, our solutions will be of the form: eigenvector * $ e^{\text{eigenvalue} * x}$.

Let's find the eigenvalues of this matrix. $$\begin{vmatrix}3-\lambda & 0 & 1\\9 & -1-\lambda & 2 \\-9 & 4 & -1-\lambda \end{vmatrix} = 0$$ $$(3-\lambda)\left((\lambda + 1)^2 + 1\right) = 0$$ When $\lambda = 3$, we get $$\begin{bmatrix}0 & 0 & 1\\9 & -4 & 2 \\-9 & 4 & -4 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0\to \begin{bmatrix}0 & 0 & 1\\9 & -4 & 0 \\-9 & 4 & 0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0 \to \begin{bmatrix}0 & 0 & 1\\9 & -4 & 0 \\0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0 $$ $$x_3 = 0, 9x_1 = 4x_2 \to \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}x_1\\ \frac{9}{4}x_1\\0\end{bmatrix} \to C\begin{bmatrix}4\\9\\0\end{bmatrix}$$ So, that specific solution becomes $\begin{bmatrix}4\\9\\0\end{bmatrix}e^{3x}$.

Problem: Imaginary Numbers

Continuing with the current question, we potentially have a problem. Our other eigenvalues are $-1 \pm i$. Let's start with $-1 + i$, and we'll see that we don't even need to do $-1-i$.Let's substract from the diagonal like a normal eigenvalue. We get $$\begin{bmatrix}4 - i & 0 & 1\\9 & -i & 2\\-9 & 4 & -i\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$$ We could solve this, or we could use a trick that the book teaches. We first let $x_1 = 1$. This is okay, because the real value of $x_1$ is any multiple of that. That means $x_3 = -4 + i$. Finally, with some arithmetic, you get $x_2 = 2-i$.

We get this as our answer $$\begin{bmatrix}1\\2-i\\-4 + i\end{bmatrix}e^{(-1 + i)t} \to \left(\begin{bmatrix}1\\2\\-4\end{bmatrix} + i\begin{bmatrix}0\\-1\\1\end{bmatrix}\right)e^{-t}\left(\cos(t) + i\sin(t)\right)$$. $$x(t) = e^{-t}\left(\begin{bmatrix}1\\2\\-4\end{bmatrix}\cos(t) - \begin{bmatrix}0\\-1\\1\end{bmatrix}\sin(t)\right) + ie^{-t}\left(\begin{bmatrix}0\\-1\\1\end{bmatrix}\cos(t) + \begin{bmatrix}1\\2\\-4\end{bmatrix}\sin(t)\right)$$ Now, we split this into two solutions $$x_2 = e^{-t}\begin{bmatrix}\cos(t)\\2\cos(t) + \sin(t)\\-4\cos(t) - \sin(t)\end{bmatrix}, x_3 = e^{-t}\begin{bmatrix}\sin(t)\\-\cos(t) + 2\sin(t)\\ \cos(t) + 4\sin(t)\end{bmatrix}$$

Problem: Nonhomogeneous

Let's say we have this problem $$x' = \begin{bmatrix}4 & 2 \\ 3 & -1\end{bmatrix}x + \begin{bmatrix}0\\e^{-2t}\end{bmatrix}$$ Using the method we learned before, we find the solution to the homogeneous equation first. That equals $$c_1\begin{bmatrix}1 \\ -3\end{bmatrix}e^{-2t} + c_2\begin{bmatrix}2 \\ 1\end{bmatrix}e^{5t}$$ In our previous problems, we would have originally guessed $\vec{a}e^{-2t}$ then changed it to $te^{-2t}$ because of the repeat root. In these types of problems, we change our guess to $x_p = Ate^{-2t} + Be^{-2t}$. $x'_p = (A-2B)e^{-2t} - 2Ate^{-2t}$. $$\begin{bmatrix}a_1 - 2b_1 - 2a_1t\\a_2 - 2b_2 -2a_2t\end{bmatrix} = \begin{bmatrix}4 & 2\\3 & -1\end{bmatrix}\begin{bmatrix}a_1t + b_1\\a_2t + b_2\end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}(4a_1 + 2a_2)t + 4b_1 + 2b_2\\(3a_1 - a_2)t + 3b_1 - b_2 + 1\end{bmatrix}$$ Now, we have to solve this and figure out what the a's and b's are. Here are the equations $$3a_1 + a_2 = 0, 6b_1 + 2b_2 = a_1, 3b_1 + b_2 = a_2 - 1$$ We use the last two equations to get $$ A = \begin{bmatrix}\frac{-2}{7}\\ \frac{6}{7}\end{bmatrix}$$ We can't get both Bs, but that's okay, because it absorbs into the previous $e^{-2t}$ expression. We get $\begin{bmatrix}b_1\\-3b_1 - \frac{1}{7}\end{bmatrix}e^{-2t}$ So, for our final answer, we get $$x = c_1\begin{bmatrix}1\\-3\end{bmatrix}e^{-2t} + c_2\begin{bmatrix}2\\1\end{bmatrix}e^{5t} + \begin{bmatrix}\frac{-2}{7}\\ \frac{6}{7}\end{bmatrix}te^{-2t} + \begin{bmatrix}0 \\ -\frac{1}{7}\end{bmatrix}e^{-2t}$$

Problem: Repeat Roots

Let's say we have: $$\vec{x}' = \begin{bmatrix}1 & -1\\4 & -3\end{bmatrix}\vec{x}$$ The only eigenvalue is -1. The only eigenvector is $\begin{bmatrix}1\\2\end{bmatrix}$. Here we have a problem. We know one solution: $\vec{x_1} = \begin{bmatrix}1\\2\end{bmatrix}e^{-t}$. We want two equations, however. So, what can we do? For $x_2$, we guess $c_1\left(\vec{v_1}te^{-t} + \vec{v_2}e^{-t}\right)$, where $\vec{v_1}$ is the first eigenvector. Here is the equation to get $v_2$. The proof is below. $$\left(A - (\lambda{}I)\right)\vec{v_2} = \vec{v_1} \text{ and } \left(A - (\lambda{}I)\right)\vec{v_1} = \vec{0} $$ Show/Hide Proof

$$x_2 = v_1te^{\lambda t} + v_2e^{\lambda t}$$ $$x_2' = Ax$$ $$v_1e^{\lambda t} + \lambda v_1te^{\lambda t} + \lambda v_2 e^{\lambda t} = Av_1te^{\lambda t} + Av_2e^{\lambda t}$$ Note, $Av_1 = \lambda v_1$, because $v_1$ is an eigenvector and $\lambda$ is the corresponding eigenvector. We can't do this with $\vec{v_2}$. $$v_1e^{\lambda t} + \lambda v_1te^{\lambda t} + \lambda v_2 e^{\lambda t} = \lambda v_1te^{\lambda t} + Av_2e^{\lambda t}$$ this with $\vec{v_2}$. $$v_1e^{\lambda t} + \lambda v_2 e^{\lambda t} = Av_2e^{\lambda t}$$ We now put the $v_2$ on the right side. $$v_1e^{\lambda t} = \left(A - \lambda I\right)v_2e^{\lambda t}$$ We divide through by $e^{\lambda t}$ $$\left(A - \lambda I\right)v_2 = v_1$$

It's not too hard, just write out $x'$ and solve. As an extension, we find that $\left(A - \lambda I\right)^2\vec{v_2} = \vec{0}$. We do that for this problem. $$\begin{bmatrix}2 & -1 \\ 4& -2\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}1\\2\end{bmatrix}$$ $$\vec{v_2} = c_1\begin{bmatrix}1\\2\end{bmatrix} + \begin{bmatrix}0\\-1\end{bmatrix}$$ The repeated $\begin{bmatrix}1\\2\end{bmatrix}$ may be absorbed. Our final solution becomes: $$c_1\begin{bmatrix}1\\2\end{bmatrix}e^{-t} + c_2\left(\begin{bmatrix}1\\2\end{bmatrix}te^{-t} + \left(c_3\begin{bmatrix}1\\2\end{bmatrix} + \begin{bmatrix}0\\-1\end{bmatrix}\right)e^{-t}\right)$$ Because we get a $(c_1 + c_2*c_3)\begin{bmatrix}1\\2\end{bmatrix}e^{-t}$. Those can be combined into one constant. Our real final answer is: $$\boxed{c_1\begin{bmatrix}1\\2\end{bmatrix}e^{-t} + c_2\left(\begin{bmatrix}1\\2\end{bmatrix}te^{-t} + \begin{bmatrix}0\\-1\end{bmatrix}e^{-t}\right)}$$