Question

Given the linear differential system
x1'=5x1 - 3x2
x2'=2x1

Let u=(e^2t, e^2t)^T and v=(3e^3t, 2e^3t)^T

Show that u, v are a fundamental set of solutions of the system.

Answer 1

Given the linear differential system.
\[x'=\left[\begin{matrix}5 & -3 \\ 2 & 0\end{matrix}\right]x\]

Answer 2

find \(\det(A- \lambda I)\)

Answer 3

what is A... the matrix from x', or the matrix from u and v?

Answer 4

A=the matrix you wrote above

Answer 5

(the system matrix)

Answer 6

Yes, What electrokid said.

Answer 7

the idea is, to find the eigen values of the system matrix

Answer 8

and the corresponding eigen vectors.

Answer 9

okay, that can be done... but is there another way. Eigen values and vectors was about 6 chapters ago, and there is no mention of that in this section.

There is one example in the book, and it just says given that
u=(t^3, 3t^2) and v=(t^-1, -t^-2)

That they form the fundamental set of solutions for
\[x'=\left[\begin{matrix}0 & 1 \\ 3/t ^{2} & 1/t\end{matrix}\right]\]

I just don't understand how he gets this matrix. Because it doesn't say anywhere about the eigenvalues/vectors. I have no problem doing it, I just want to make sure I am not missing what I am supposed to be understanding

Answer 10

those seem like eigen vectors!

Answer 11

okay, so when I do calculate the eigen vectors. What do those have to do with u and v?

Answer 12

fundamental solutions = e^(eigenvalue*t) * [eigen vector]^T

Answer 13

which will be u and v, so there is only one fundamental set of solutions

Answer 14

no.. since you have eigen vectors as the non-zero scalar mmultiples, you can have a non-zero scalar multiples of 'u' and 'v'
given some initial conditions, you can solve for those constants

Answer 15

okay, i will try it out, thanks for the help.

Answer 16

yw