system of ODE

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Apr 24th 2006, 10:51 PM
braddy

system of ODE

Hi, Please can someone help me on how to do this exercise.

Give a fundamental matrix for the system:
{x'(t)=-y(t)
{y'(t)=20x(t)-4y(t)

the solution is like:
{v1(t)=e^(2t)*cos(4t)[1;4]+e^(2t)*sin(4t)[-1;-2], v2(t)=e^(2t)*cos(4t)[1;4]+e^(2t)*sin(4t)[0;-4]}

[1;4].......are colunm vectors.
IT is just a form sample .since it is a multiple choice question, I just took one solution to show you what it might look like.

Thank you,
B
Apr 24th 2006, 11:51 PM
CaptainBlack

Quote:

Originally Posted by braddy

Hi, Please can someone help me on how to do this exercise.

Give a fundamental matrix for the system:
{x'(t)=-y(t)
{y'(t)=20x(t)-4y(t)

the solution is like:
{v1(t)=e^(2t)*cos(4t)[1;4]+e^(2t)*sin(4t)[-1;-2], v2(t)=e^(2t)*cos(4t)[1;4]+e^(2t)*sin(4t)[0;-4]}

[1;4].......are colunm vectors.
IT is just a form sample .since it is a multiple choice question, I just took one solution to show you what it might look like.

Thank you,
B

Not sure this is what you need but the system is equivalent to:

$ X'(t)=\left[ \begin{array}{cc}0 & -1\\20 & -4\end{array} \right]X(t) $

where $X(t)$ and $X'(t)$ are column vectors.

RonL
Apr 25th 2006, 06:09 AM
braddy

Quote:

Originally Posted by CaptainBlack

Not sure this is what you need but the system is equivalent to:

$ X'(t)=\left[ \begin{array}{cc}0 & -1\\20 & -4\end{array} \right]X(t) $

where $X(t)$ and $X'(t)$ are column vectors.

RonL

Ok I have trouble converting it into a system .
Yes this is it . I know how to solve it.
Thank you captain
B
Apr 25th 2006, 06:20 AM
CaptainBlack

Quote:

Originally Posted by braddy

Hi, Please can someone help me on how to do this exercise.

Give a fundamental matrix for the system:
{x'(t)=-y(t)
{y'(t)=20x(t)-4y(t)

the solution is like:
{v1(t)=e^(2t)*cos(4t)[1;4]+e^(2t)*sin(4t)[-1;-2], v2(t)=e^(2t)*cos(4t)[1;4]+e^(2t)*sin(4t)[0;-4]}

[1;4].......are colunm vectors.
IT is just a form sample .since it is a multiple choice question, I just took one solution to show you what it might look like.

Thank you,
B

It is of course also equivalent to the 2-nd order ODE:

$ \ddot y(t)=-20y(t)+4\dot y(t) $

or:

$ \ddot y(t)-4\dot y(t)+20y(t)=0 $

RonL
Apr 25th 2006, 05:50 PM
Rebesques

Quote:

Yes this is it . I know how to solve it.

Don't tell me you 'll go for e^{At}... :confused: :o