Question

Find two linearly independent solutions of....

see below.

Answer 1

\[x'=\left[\begin{matrix}0 & 1 \\ -4/t ^{2} & 3/t\end{matrix}\right]x\]

Answer 2

\[\det(A-\lambda I)=\lambda ^{2}-3/t \lambda +4/t ^{2}\]

Answer 3

the equivalent second order equation is

\[y''-3/ty'+4/t ^{2}y=0\]

so therefore \[y(t)=c _{1}t ^{2}+c _{2}t ^{2}\log(t)\]

and the fundamental solution set is

\[{t ^{2} , t ^{2} \log(t)}\]

but how can I find independent solutions, solving for the vectors v1 and v2???

Answer 4

@wio @zepdrix

Answer 5

Sorry, I'm a bit confused.

Answer 6

So we already have a system of equations, and on top of it, there is a \(t\) involved?

Answer 7

Are we differentiating with respect to \(t\)?

Answer 8

Yes, x'=Ax is given to us, and I am supposed to find two solutions of that... the rest of the info I just figured out from what was given.

Answer 9

I have an example with the solution for a problem just like that. But I don't see what was differentiated if anything

Answer 10

When you say \(x'\) you are saying it's a derivative. A derivative is always the result of differentiating with respect to some variable.

Answer 11

Basically, there is important context missing that is probably make explicit in the book somewhere else.

Answer 12

well these are just like the other problems that you helped me with before, when we solved for lambda, found the eigenvectors, and multiplied that by the fundamental solution set, but in all the other case, the matrix was just constants... and the fundamental solution set was just {e^lambda1t, e^lambda2t} times the eigenvector.

Answer 13

but I cant get the eigenvector for these problems with t in it.

Answer 14

would it help at all if I posted the question that i do have the solution for and maybe can see what I mean, and maybe that would give some ideas.

Answer 15

Hmmm, so the issue is converting your second order solution back into matrix form?

Answer 16

for these problems yes, i can get the fundamental solution set, but I can not get the vectors v1 and v2 that you multiply by the solution set in order to get the matrix V(t) containing the two solutions.

Answer 17

Are those vectors are going to depend on \(t\)?

Answer 18

What I can say is this.... we let \(y=x_2\) and \(y'=x_1\)

Answer 19

whoops. I mean \(y=x_1,y'=x_2\). We used this fact to make it second order.

Answer 20

and what do I do with that to get my vectors?

Answer 21

\[
x_1'=0x_1+1x_2 = y'\\
x_2'=(-4/t^2)x_1+(3/t)x_2 = (-4/t^2)y+(3/t)y'

\]

Answer 22

I'm not sure what the vectors or for, but this certainly is a matrix, right?

Answer 23

\[
y(t)=c _{1}t ^{2}+c _{2}t ^{2}\log(t)\\
y'(t)=c _{1}2t+c _{2}[2t\log(t)+t]
\]

Answer 24

If you want to solve for the constants, you need initial conditions.

Answer 25

okay, i follow you, let me try that real quick with the problem that i know the solution too and see if it works out.

Answer 26

well that worked out, all i have to do was take the derivative of y(t) and use the two equations... so in this problem my two solutions will be.

\[x _{1}(t)=\left(\begin{matrix}t ^{2} \\ 2t\end{matrix}\right)\]

\[x _{2}(t)=\left(\begin{matrix}t ^{2}\log t \\ 2t logt + t\end{matrix}\right)\]

Answer 27

I appreciate your help, I made that way harder than it needed to be.

Answer 28

Well, I wouldn't have been able to do it on my own. All I really could do was speculate a bit on the nature of the matrix.

Answer 29

well it works, and it worked with the problem I know the solution, so all ended well. Thank you for the help.