# Math Help - system of ODE

1. ## 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

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

3. 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

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

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

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