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Fundamental matrix

  • Thread starter soopo
  • Start date
  • #1
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Homework Statement



Show that the vectors
[tex]\overline{x}(t) =
\begin{bmatrix}
e^t
\\ -t
\end{bmatrix}
[/tex]

and

[tex]
\overline{y}(t) =
\begin{bmatrix}
t
\\ e^(-t)
\end{bmatrix}[/tex]
are solutions for

[tex]\overline{x}' = \frac {1} {1+t^2}
\begin{bmatrix}
1+t & e^t(1-t)
\\ -e^-(t) (1+t) & t-1
\end{bmatrix}
\overline{x}
.[/tex]

The last row should be a 2x2 matrix. The x and y are vectors.



The Attempt at a Solution



I feel that it would be helpful to have fundamental matrix here: [itex]\Phi(t,s)[/itex].
 

Answers and Replies

  • #2
statdad
Homework Helper
1,495
35
Perhaps I'm missing something: why don't you simply substitute the given vectors into the matrix system and see if things balance?
 
  • #3
33,646
5,313
Perhaps I'm missing something: why don't you simply substitute the given vectors into the matrix system and see if things balance?
You're not missing anything at all. soopo should take x(t), find x'(t), and then observe that these vectors make the differential equation identically true.

Then do exactly the same thing for y(t) and y'(t).
 
  • #4
225
0
Perhaps I'm missing something: why don't you simply substitute the given vectors into the matrix system and see if things balance?
I get [itex] x' = (e^t, -1) [/itex] by putting x to x' and [itex] x' = (1,-e^t)[/itex] by putting y to x'.
This gives me [itex]t = -t [/itex] that is [itex] 1 = -1[/itex] which is false.
 
Last edited:
  • #5
33,646
5,313
I get [itex] x' = (e^t, -1) [/itex] by putting x to x' and [itex] x' = (e^t, -1)[/itex] by putting y to x'.
This gives me [itex]t = -t [/itex] that is [itex] 1 = -1[/itex] which is false.
What do you mean "by putting y to x'"? You have what amounts to two problems: showing that x(t) = (et, -t) is a solution of the differential equation and showing that y(t) = (t, e-t) is a solution.

I have done the first part and have confirmed that x(t) is a solution.
 
  • #6
225
0
What do you mean "by putting y to x'"? You have what amounts to two problems: showing that x(t) = (et, -t) is a solution of the differential equation and showing that y(t) = (t, e-t) is a solution.

I have done the first part and have confirmed that x(t) is a solution.
I had a typo in my reply.
I mean by "putting y to x" that you make y equals x in the last statement for x'.

I differentiate the x by getting the same x' which I get by plugging in the x for the statement x'.

How can you show that x(t) = (et, -t) is a solution of the differential equation?
 
Last edited:
  • #7
33,646
5,313
What you are calling the "last statement for x'" is the differential equation. You are trying to show that the vector functions x(t) and y(t) are solutions of this differential equation.

Are you trying to show that x(t) = y(t) and that x'(t) = y'(t)? If so, that's obviously not true, and that's not what this problem is about.
 
  • #8
statdad
Homework Helper
1,495
35
"Perhaps I'm missing something..."

I hadn't had my morning coffee when I posted, so I needed to hedge my bets. :}

soopo: differentiate the vector [tex] x [/tex], and perform the matrix multiplication using tex] x [/tex] on the right. You should get the sam results - this shows your given vector is a solution. Do the same steps for [tex] y [/tex].
 
  • #9
225
0
"Perhaps I'm missing something..."

I hadn't had my morning coffee when I posted, so I needed to hedge my bets. :}

soopo: differentiate the vector [tex] x [/tex], and perform the matrix multiplication using tex] x [/tex] on the right. You should get the sam results - this shows your given vector is a solution. Do the same steps for [tex] y [/tex].
My y -value [itex] obtained by differentiation is different from the one obtained by substitution that is
[tex] (1, -e^(-t)) != (t^2 + 1, -2e^(-t))[/tex]
 

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