Question

Linear Differential Equation

Answer 1

Can someone help my with this exercise?

Find all functions x(t), y(t) satisfying
\[x'(t)=y(t)-x(t)\]
\[y'(t)=3x(t)-3y(t)\]
Find the function pair of functions satisfying x(0)=y(0)=1/2

Answer 2

\[{\bf{x}'}=\begin{pmatrix}-1&1\\3&-3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\]
Find your eigenvalues first:
\[\begin{vmatrix}-1-\lambda&1\\3&-3-\lambda\end{vmatrix}=(1+\lambda)(3+\lambda)-3=0\\
\lambda^2+4\lambda=0\\
\lambda_1=0,~\lambda_2=-4\]
Do you know how to find eigenvectors?

Answer 3

Yes

Answer 4

So what do you get?

Answer 5

((-1, 3), (1,1))

Answer 6

Right. The form of the solution would then by
\[{\bf x}=C_1\eta_1e^{\lambda_1 t}+C_2\eta_2 e^{\lambda_2 t}\]
where \(\eta_{1,2}\) are the respective eigenvectors for \(\lambda_{1,2}\).

Answer 7

Particularly,
\[{\bf x}=C_1\begin{pmatrix}1\\1\end{pmatrix}+C_2\begin{pmatrix}-1\\3\end{pmatrix}e^{-4t}\]
Plug in your initial values and solve for the constants, and you're done.

Answer 8

Setting up your equation, you get
\[{\bf x}(0)=\begin{pmatrix}x(0)\\y(0)\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\\\\\frac{1}{2}\end{pmatrix}\]
and so you get the following system of equations:
\[\begin{cases}C_1-C_2=\frac{1}{2}\\C_1+3C_2=\frac{1}{2}\end{cases}\]

Answer 9

Thank you. My textbook solve the equation another way. Can you see the different?

Answer 10

I remember learning to solve equations that way, but I don't remember the details, sorry. I had trouble understanding this method the first time around and never really bothered to review it since.

Answer 11

If you're having trouble following the reasoning, there are quite a few videos of lectures on Youtube you can watch, hopefully with examples.

Answer 12

I will not blame you. :)

Answer 13

Thank you.

Answer 14

You're welcome! At the least, you have a known solution, so you'll know whether or not you've done something wrong if you were to try this method.