3rd time I'm asking you guys this question.

[Jamin2112]

Homework Statement

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

(a) Compute the Wronskian of 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 homogenous differential equations satisfied by x⁽¹⁾ and x⁽²⁾?
(d) Find this system of equations and verify the conclusion of part (c).

Homework Equations

Wronskian is a fancy word for the determinate of a set of solutions.
A set of vectors is linearly independent <---> Determinate ≠ 0.

The Attempt at a Solution

Wronskian = |x⁽¹⁾, x⁽²⁾| = t²

t² = 0 ---> No solution; the system is not linearly independent. So our interval is t ≠ 0.

Parts (c) and (d) don't make sense to me.

Making x⁽¹⁾, x⁽²⁾ a system of homogenous equations would mean C₁x⁽¹⁾ + C₂x[/Bb]⁽²⁾ = 0.

The answer in the back of the book says, "At least one coefficient must be discontinuous at t=0." How can a coefficient, just a number, be discontinuous?

Show me your mathleticism on this one, folks.

 

[Gregg]

I'm out of my depth by even looking at it. But if you have concluded that the interval to be t\ne 0 can you say something about continuity at t=0?

 

[HallsofIvy]

Homework Statement

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

(a) Compute the Wronskian of 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 homogenous differential equations satisfied by x⁽¹⁾ and x⁽²⁾?
(d) Find this system of equations and verify the conclusion of part (c).

Homework Equations

Wronskian is a fancy word for the determinate of a set of solutions.

Specifically, the Wronskian of f(x) and g(x) is
\left|\begin{array}{cc}f &amp; g \\ f&#039; &amp; g&#039;\end{array}\right|

A set of vectors is linearly independent <---> Determinate ≠ 0.

The Attempt at a Solution

Wronskian = |x⁽¹⁾, x⁽²⁾| = t²

t² = 0 ---> No solution; the system is not linearly independent. So our interval is t ≠ 0.

Well, "t\ne 0" isn't an interval. Your two intervals are t< 0 and t> 0.

Parts (c) and (d) don't make sense to me.

Making x⁽¹⁾, x⁽²⁾ a system of homogenous equations would mean C₁x⁽¹⁾ + C₂x[/Bb]⁽²⁾ = 0.

The answer in the back of the book says, "At least one coefficient must be discontinuous at t=0." How can a coefficient, just a number, be discontinuous?

Why do you say that the coefficients must be numbers? We are talking about the coefficients of the dependent function and its derivatives. Many ("almost all") linear differential equations have variable coefficients.

Show me your mathleticism on this one, folks.

We seek a differential equation of the form
\frac{dX}{dt}= \begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}X

that is satisfied by x1 and x2. Here, dx1/dt= (1, 0) and dx2/dt= (2t, 2).

We must have
\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}\begin{bmatrix}t \\ 1\end{bmatrix}
and
\begin{bmatrix}2t \\ 2\end{bmatrix}= \begin{bmatrix} a &amp; b \\ c &amp; d\end{bmatrix}\begin{bmatrix}t^2 \\ 2t\end{bmatrix}

That means we must have at+ b= 1, ct+ d= 0, at^2+ 2bt= 2t, and ct^2+ 2dt= 2.
From at+ b= 1, b= 1- at. Putting that into at^2+ 2bt= 2t gives at^2+ 2(1- at)t= at^2+ 2t- 2at^2= 2t which reduces to at^2= 0[/tex] or a= 0 and then b= 1.<br /> <br /> From ct+ d= 0, d= -ct. Putting that into ct^2+ 2dt= 2 gives ct^2+ 2(-ct)t= -ct^2= 2 so that c= -2/t^2 and then d= 2/t.<br /> <br /> The given functions satisfy the equation <br /> \frac{\begin{bmatrix}x(t) \\ y(t)\end{bmatrix}}{dt}= \begin{bmatrix}0 &amp;amp; 1 \\ -\frac{2}{t^2} &amp;amp; \frac{2}{t}\end{bmatrix}\begin{bmatrix}x(t) \\ y(t)\end{bmatrix}

 

[lanedance]

so you're using Boyce & DiPrima? just consulted my old text for a refresher & saw the problem in section 7.4 (though mines 5th edition)

So anyway, a) & b) look ok, and I think where you're going wrong in c) & d) is the understanding of the "coefficients" referred to... doing d) first may help... though here's an explanantion I hope helps.

The solution of a homgenous ODE will be composed of a linear combination of several functions with arbitrary constants.
c_1x_1(t)+ c_2x_2(t) +...
When you apply boundary/initial values to the problem the constants will only be allowed certain values to satisfy - i don't think these are the coefficients referred to in the question.

Going back to the start of section 7.4, the usual way to write a system of first order homogenous DEs takes the form
\textbf{x}&#039;(t) = \textbf{P}(t)\textbf{x}(t)

each of \textbf{x}^{(1)}, \textbf{x}^{(2)} will be a solution of the above equation

note P(t), the 2x2 coeffcient matrix, is a function of t and generally referred to as the coefficients of the system of DEs.

Going back to the question:

(c) What conclusion can be drawn about the coefficients in the system of homogenous differential equations satisfied by x⁽¹⁾ and x⁽²⁾?
(d) Find this system of equations and verify the conclusion of part (c).

going back to Thm 7.1.2 if the coefficients of a system of DEs are continuous on an open interval, then there exists unique solutions on that interval.

The fact your Wronskian disappears at t=0 shows the solutions are not unqiue for all t on an open interval containing t = 0. So what can you then say about the coefficients..?

 

[lanedance]

didn't see Halls post until after (could have saved me some work), added some edits above to align

 

[Jamin2112]

Specifically, the Wronskian of f(x) and g(x) is
\left|\begin{array}{cc}f &amp; g \\ f&#039; &amp; g&#039;\end{array}\right|


Well, "t\ne 0" isn't an interval. Your two intervals are t< 0 and t> 0.


Why do you say that the coefficients must be numbers? We are talking about the coefficients of the dependent function and its derivatives. Many ("almost all") linear differential equations have variable coefficients.



We seek a differential equation of the form
\frac{dX}{dt}= \begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}X

that is satisfied by x1 and x2. Here, dx1/dt= (1, 0) and dx2/dt= (2t, 2).

We must have
\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}a &amp; b \\ c &amp; d\end{bmatrix}\begin{bmatrix}t \\ 1\end{bmatrix}
and
\begin{bmatrix}2t \\ 2\end{bmatrix}= \begin{bmatrix} a &amp; b \\ c &amp; d\end{bmatrix}\begin{bmatrix}t^2 \\ 2t\end{bmatrix}

That means we must have at+ b= 1, ct+ d= 0, at^2+ 2bt= 2t, and ct^2+ 2dt= 2.
From at+ b= 1, b= 1- at. Putting that into at^2+ 2bt= 2t gives at^2+ 2(1- at)t= at^2+ 2t- 2at^2= 2t which reduces to at^2= 0[/tex] or a= 0 and then b= 1.<br /> <br /> From ct+ d= 0, d= -ct. Putting that into ct^2+ 2dt= 2 gives ct^2+ 2(-ct)t= -ct^2= 2 so that c= -2/t^2 and then d= 2/t.<br /> <br /> The given functions satisfy the equation <br /> \frac{\begin{bmatrix}x(t) \\ y(t)\end{bmatrix}}{dt}= \begin{bmatrix}0 &amp;amp; 1 \\ -\frac{2}{t^2} &amp;amp; \frac{2}{t}\end{bmatrix}\begin{bmatrix}x(t) \\ y(t)\end{bmatrix}

<br /> <br /> <br /> I see. So I assume there&#039;s some 2x2 Matrix <b>A</b> that can be multiplied by each of these solution vectors to get the corresponding derivative vectors (which are simply vectors in which one takes the derivative of each component). &quot;Coefficient&quot; just means an entry in the matrix. <br /> <br /> Thanks a million, sir!