Linear ODE Systems in Numerical Methods.

[arizonian]

I feel so embarrased asking this question, but this is the place to get answers.

I have a 2nd order ODE with a forcing function that needs to be manipulated and put into a matrix for a numerical method solution, ie Matlab. My question is: Is the matrix composed of a particular solution in the top row and a homogenous solution in the bottom row? Does this satisfy the requirement for two equations? Maybe I should say equation, not solution.

My work:

m d^2x/dt^2 + c dx/dt + kx = 0

d^2x/dt^2 = dx/dt

Substituting y2 for d^2t/dx^2 and y1 for dx/dt, and realizing that y2 is the derivative of y1, I end up with, in matrix form:

(I am using periods to hold the spacing)

[-1/k......-c/mk]..[y2]...=[x2]
[1...............-1]..[y1]...=[x1]

Thank you

Bill

On edit, I realized I forgot the signs in the first equation.
On second edit, I changed the lower equation to simplify what I was after.

[arizonian]

Any body care to comment?

Bill

[oookhc]

The way you describe the problem is a little confusing. Let me try to paraphrase.
Your original problem is:
m \frac{d^2 x}{d t^2} + c \frac{dx}{dt} + kx = 0
m \frac{d^2 x}{d t^2} - \frac{dx}{dt} = 0

And you would like to transfer it to matrix form, right ?
You might do something as follows:

y_{1}=x
y_{2}= \frac{dx}{dt}
And let
X = \left(\begin{array}{c} y_{1} \\ y_{2} \end{array}\right)
Y = \left(\begin{array}{c}\frac{d y_{1}}{dt} \\ \frac{dy_{2}}{dt} \end{array}\right)

And you might turn this problem into

\left(\begin{array}{cc} c & m \\ -1 & m \end{array}\right) Y=
\left(\begin{array}{cc} -k & 0 \\ 0 & 00 \end{array}\right)X


Is this what you are trying to ask ?

Probably you can just start from here and check the other websites
to solve this question in martrix form. You might as well just take a look
at

http://www.ScienceOxygen.com/math409.html

It might not solve your question directly. But it is with a lot of links
on differential equation. You could start from there to collect some information...

[saltydog]

I think the original question is a bit confussing. This is how I'd present converting a high-order ODE to a system of first-order ODEs:



\frac {d^2y} {dx^2} + \ln{y} = yx


To convert this to a system of ODEs,

let:


z[x]=\frac{dy} {dx}


Then:


\frac{dy} {dx} = z



\frac {dz} {dx} =yx-\ln{y}


You can use the same method for higher-order ODEs, just assigning different variables to each higher-order derivative. I used Mathematica to solve this system with initial conditions:


y[0]=1



y'[0]=1


I attached a plot of the solution which you can find in "additional options".

SD

[saltydog]

Oh yea, I thought the talking dictionary on line was cool, but LaTeX rocks!

SD