A Homogeneous Linear System w/ Constant Coefficients

Jamin2112
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I'll make this post short.

The problem just asks me to something in the form x'=Ax (A is a 2x2 of constants) and then describe the behavior of the solution as t approaches infinity.

My solution is x=C1e-2t(2/3 1)T + C2e-t(1 1)T.

Since both vectors are multiplied by 1/et, my solution just goes to (0 0)T as t->∞, right?
 
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Hi Jamin2112! :smile:
Jamin2112 said:
My solution is x=C1e-2t(2/3 1)T + C2e-t(1 1)T.

Since both vectors are multiplied by 1/et, my solution just goes to (0 0)T as t->∞, right?

Yes …

but from which direction? :wink:

(and if your solution is for x, not xT, why is it a transpose? :confused:)
 
tiny-tim said:
Hi Jamin2112! :smile:


Yes …

but from which direction? :wink:

(and if your solution is for x, not xT, why is it a transpose? :confused:)


Should indeed be xT

From the sketch I drew, it approaches the origin along y=x in quadrant 1 and goes away from the origin along y=x in quadrant 3. Correct?
 
Yes, it approaches the origin along y=x. :smile:

But what do you mean by "goes away from the origin" …

isn't there only one way of getting to t = ∞ ? :confused:
(and what if C2 = 0 ? :wink:)
 
tiny-tim said:
Yes, it approaches the origin along y=x. :smile:

But what do you mean by "goes away from the origin" …

isn't there only one way of getting to t = ∞ ? :confused:
(and what if C2 = 0 ? :wink:)

The equation was x'=[1 -2; 3 -4]x (wrote the matrix MATLAB-style)

So at x=(1 1)T, x'=(-1 -1)T, and so on ...

And at x=(-1 -1)T, x'=(1 1)T, and so on ...

See what I mean?
 
In general, I'm confused about these questions about what x(t) approaches as t approaches infinity.

For a different problem on my homework, I got a solution x= 2e3t(1 5)T -e-t(1 1)T. Obviously, the first part, 2e3t(1 5)T, will dominate as t --> ∞. Does that mean the x approaches (∞ 5∞)T?
 
Jamin2112 said:
In general, I'm confused about these questions about what x(t) approaches as t approaches infinity.

For a different problem on my homework, I got a solution x= 2e3t(1 5)T -e-t(1 1)T. Obviously, the first part, 2e3t(1 5)T, will dominate as t --> ∞. Does that mean the x approaches (∞ 5∞)T?

What happens as t approaches ∞ is the same for matrices as it is for ordinary numbers …

so yes in your example, 2e3t(1 5)T, will dominate, and that approaches infinity along the (1 5)T direction. :smile:
 
tiny-tim said:
What happens as t approaches ∞ is the same for matrices as it is for ordinary numbers …

so yes in your example, 2e3t(1 5)T, will dominate, and that approaches infinity along the (1 5)T direction. :smile:

I see. Can I ask one more question?
 
uhhh? :redface: depends what it is! :biggrin:
 
  • #10
tiny-tim said:
uhhh? :redface: depends what it is! :biggrin:

Consider the vectors x(1)(t)=(t 1)T and x(2)(t)=(t2 2t)T.

(a) Compute the Wronskian x(1) and x(2).
(b) In what intervals are x(1) and x(2) linearly independent?
(c) What conclusion can be drawn about the coefficients in the system of homogeneous differential equations satisfied by x(1) and x(2)?
(d) Find this system of equations and verify the conclusions of part (c).

------------------------*snip*----------------------

Wronskian ≠ 0 <==> system has a solution

W|(t 1)T, (t2 2t)T|= t2, which only equals zero when t=0.

Making x(1), x(2) into a system of homogenous differential equations would look like

C1(t 1)T + C2(t2 2t)T=(0 0)T.

I don't understand what the point of (c) is. Where are they going with this? Please explain in as much or as little detail as possible parts (c) and (d). I'm ready to learn.
 
  • #11
Hi Jamin2112! :smile:

(just got up :zzz: …)

Sorry, Wronskians are not my field :redface:

you'd better start another thread on this one. :smile:
 
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