Mathwizurd.com is created by David Witten, a mathematics and computer science student at Stanford University. For more information, see the "About" page.

The Eigenvalue Method

Homogeneous

MathJax TeX Test Page 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

MathJax TeX Test Page 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

MathJax TeX Test Page 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. $$\left(A - (\lambda{}I)\right)\vec{v_2} = \vec{v_1} \text{ and } \left(A - (\lambda{}I)\right)\vec{v_1} = \vec{0} $$

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: $$\boxed{c_1\begin{bmatrix}1\\2\end{bmatrix}e^{-t} + c_2\left(\begin{bmatrix}1\\2\end{bmatrix}te^{-t} + \begin{bmatrix}0\\-1\end{bmatrix}e^{-t}\right)}$$
David Witten

Laplace Transforms

Reduction of Order