Question

Find the general solution of x'=(3, 2, -2, -2)x. (this is matrix, 3 and 2 on the left, -2 and -2 on the right.)

Answer 1

\[\left[\begin{matrix}3&2\\-2&-2\end{matrix}\right]\]
step 1: find eigenvalue: \[\left[\begin{matrix}3-\lambda&2\\-2&-2-\lambda\end{matrix}\right]\]characteristic equation:
\[(3-\lambda)(-2-\lambda)+4=0\]

Answer 2

How do you find the eigenvalue?

Answer 3

hey, this information comes from linear algebra course. If you didn't take that course, how can you be in differential equation class?

Answer 4

I suck at Linear Algebra. Sorry.

Answer 5

I didn't finish the first step yet.

Answer 6

Please continue.

Answer 7

so, do you know how to construct differential equation?

Answer 8

ok, from the equation above, I have the simplified form is \[\lambda^2-\lambda-2=0\\(\lambda -2)(\lambda+1)=0\\\lambda_1=2~~~and~~~ \lambda_2=-1 \]
those are eigenvalues of the matrix. From those, plug into the second matrix, which I construct above to find eigenvectors

Answer 9

For \(\lambda_1=-1\)
the matrix above (mine) becomes
\[\left[\begin{matrix}4&2\\-2 &-1\end{matrix}\right]\], so, just calculate it by
\[\left[\begin{matrix}4&2\\-2 &-1\end{matrix}\right] \left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)\]

Answer 10

so, -2a =b, if I chose a =1, so b = -2. therefore \[\left(\begin{matrix}1\\-2\end{matrix}\right)\]is eigenvector corresponds to eigenvalue -1
Can you imitate my stuff to find out the eigenvector corresponds to eigenvalue 2?
We are near final answer, girl. PRACTICE, Please

Answer 11

How did you get -2a=b?

Answer 12

from the matrix |4 2 |
| -2 -1|
you can pick the last row time with vector (a,b) that mean -2a -b=0 or -2a =b