Question

Find the general solution of the given system:

dx/dt = 3x -y
dy/dt = 9x -y

my book is saying that the eigenvalue is 0 using the det (A-Iλ)K = 0 and finding the determinant of A-Iλ gives me a complex number. Kinda lost as to where I went wrong.

Answer 1

\[{\bf x}'=\begin{pmatrix}3&-1\\9&-1\end{pmatrix}{\bf x}\]
with \({\bf x}=\begin{pmatrix}x(t)\\y(t)\end{pmatrix}\). Let's find those eigenvalues first:
\[\begin{vmatrix}3-\lambda&-1\\9&-1-\lambda\end{vmatrix}=-(3-\lambda)(1+\lambda)+9=\lambda^2-2\lambda+6=0\]
As you said, we'll get some complex roots:
\[\lambda_{1,2}=1\pm i\sqrt{5}\]
Is this what you have so far?

Answer 2

that is what I got, but the answer key is just saying that eigenvalues are 0, so λ^2 = 0

Answer 3

(here if you want to see) http://imgur.com/QeHBWWx

Answer 4

Let's not doubt ourselves because the solution manual is telling us we're wrong. They're written by humans after all.

See here: http://www.wolframalpha.com/input/?i=Eigenvalues%5B%7B%7B3%2C-1%7D%2C%7B9%2C-1%7D%7D%5D

Answer 5

alright, so afterwards, how would I deal with the complex eigenvalues?

Answer 6

When you're dealing with just one linear, constant-coefficient ODE, what do you when you have complex roots to the characteristic equation? We introduce the Euler formula, and we end up with solutions in the form of sine and cosine functions. We do the same here.

Without delving in the computation of the eigenvector just yet (I'll call them \(\vec{\eta}_1\) and \(\vec{\eta}_2\) for now), you have the general solution
\[{\bf x}=C_1\vec{\eta}_1e^{i(1+i\sqrt5)}+C_2\vec{\eta}_2e^{i(1-\sqrt5)}=\left(C_1\vec{\eta}_1\cos\sqrt5t+C_2\vec{\eta}_2\sin\sqrt5t\right)e^t\]

Answer 7

isnt the euler formula \[c _{1} e ^{\alpha t}(\sin \beta t + \cos \beta t ) ?\]

Answer 8

but even with the general solution, the complex number i won't be possible to plug in my initial matrix to find the eigen vector, right? I would need to only use my "real number" part of the expression?

Answer 9

The constants \(C_1\) and \(C_2\) are not restricted to be real numbers only, so there's no need to consider just the real part. As with the ODE \(y''+y=0\), which has the characteristic equation \(r^2+1=0\) and hence roots \(r=\pm i\), we end up with the general solution \(y=C_1\cos t+C_2\sin t\), despite the fact that the Euler formula introduces the number \(i\). The same principle is at work here.

Answer 10

well when I meant using the real number i meant for the eigenvector because, at that point, wouldn't it be impossible to find eigen vector?

Answer 11

I'll start demonstrating how to compute the eigenvectors. Consider \(\lambda_1=1+i\sqrt5\), and denote the associated eigenvector by \(\vec{\eta}_1=\begin{pmatrix}\eta_{1,1}\\\eta_{1,2}\end{pmatrix}\):
\[\begin{pmatrix}2-i\sqrt5&-1\\9&-2-i\sqrt5\end{pmatrix}\begin{pmatrix}\eta_{1,1}\\\eta_{1,2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
Multiplying the first row by \(-9\) and the second row by \(2-i\sqrt5\):
\[\begin{pmatrix}-9(2-i\sqrt5)&9\\9(2-i\sqrt5)&-9\end{pmatrix}\begin{pmatrix}\eta_{1,1}\\\eta_{1,2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
since \((-2-i\sqrt5)(2-i\sqrt5)=-9\). Now, adding the first row to the second gives \(0\) in both columns:
\[\begin{pmatrix}-9(2-i\sqrt5)&9\\0&0\end{pmatrix}\begin{pmatrix}\eta_{1,1}\\\eta_{1,2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
The first row tells us that \(-9(2-i\sqrt5)\eta_{1,1}+9\eta_{1,2}=0\), or \(\eta_{1,2}=(2-i\sqrt5)\eta_{1,1}\). If we let \(\eta_{1,1}=1\), then we have \(\eta_{1,2}=2-i\sqrt5\), so the first eigenvector could be
\[\vec{\eta}_1=\begin{pmatrix}1\\2-i\sqrt5\end{pmatrix}\]

Answer 12

ah I see, so that answer , would still apply to my regular procedure, just using the complex value and plugging in my eigen vector in my euler formula.

Answer 13

Right, getting a compact form of the final answer is a matter of algebraic manipulation.

Answer 14

well there would be a 2 part answer, right? because I would be dealing with a 2 variable matrix right?

Answer 15

Well, you still have to deal with the other eigenvector. You can go through the computation again, and you'd find that complex eigenvectors come in conjugate pairs, so that \(\vec{\eta}_2=\begin{pmatrix}1\\2+i\sqrt5\end{pmatrix}\).

So for now, you have this form of the fundamental solution (omit the constants \(C_1,C_2\)):
\[\begin{align*}{\bf x}&={\bf x}_1+{\bf x}_2\\\\
&=\begin{pmatrix}1\\2-i\sqrt5\end{pmatrix}e^{(1+i\sqrt5)t}+\begin{pmatrix}1\\2+i\sqrt5\end{pmatrix}e^{(1-i\sqrt5)t}\\\\
&=\begin{pmatrix}1\\2-i\sqrt5\end{pmatrix}e^t(\cos\sqrt5t+i\sin\sqrt5t)+\begin{pmatrix}1\\2+i\sqrt5\end{pmatrix}e^t(\cos\sqrt5t-i\sin\sqrt5t)\end{align*}\]
Let's consider the first solution alone:
\[\begin{array}{c}\begin{pmatrix}1\\2-i\sqrt5\end{pmatrix}e^t(\cos\sqrt5t+i\sin\sqrt5t)\\\\
\begin{pmatrix}\cos\sqrt5t+i\sin\sqrt5t\\(2-i\sqrt5)(\cos\sqrt5t+i\sin\sqrt5t)\end{pmatrix}e^t\\\\
\begin{pmatrix}\cos\sqrt5t+i\sin\sqrt5t\\2\cos\sqrt5t-i\sqrt5\cos\sqrt5t+2i\sin\sqrt5t+\sqrt5\sin\sqrt5t\end{pmatrix}e^t\\\\
{\bf x}_1=\begin{pmatrix}\cos\sqrt5t\\2\cos\sqrt5t+\sqrt5\sin\sqrt5t\end{pmatrix}e^t+i\begin{pmatrix}\sin\sqrt5t\\2\sin\sqrt5t-\sqrt5\cos\sqrt5t\end{pmatrix}e^t
\end{array}\]
The *first* general solution is then
\[{\bf x}_1=C_1\begin{pmatrix}\cos\sqrt5t\\2\cos\sqrt5t+\sqrt5\sin\sqrt5t\end{pmatrix}e^t+C_2\begin{pmatrix}\sin\sqrt5t\\2\sin\sqrt5t-\sqrt5\cos\sqrt5t\end{pmatrix}e^t\]

Answer 16

you didn't have to solve it, but it clears it up, you're awesome man thanks for the help!

Answer 17

There's still the other solution to find, I'll leave that to you. You're welcome!