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CaptainBlack · #4 November 27th, 2008, 10:58 PM

Quote:

Originally Posted by Rapha

Hello.

that was very helpful, thank you very much (and yes, i could get ode45 to work).

But actually i kinda screwed it(sorry, i did not exactly realize what my problem was and asked for something different), because my problem is :

$\ddot{x} = x + 2 \dot{y} - 0.9 \frac{x+0.1}{D_1} - 0.1 \frac{x-0.9}{D_2}$

$\ddot{y} = y - 2 \dot{x} - 0.9 \frac{y}{D_1} - 0.1 \frac{y}{D_2}$

with $D_1 = \sqrt{(((x+0.1)^2+y^2)^3)}$

$D_2 = \sqrt{(((x-0.9)^2+y^2)^3)}$

i tried to find the IVP 1. order:

$x_1 = x$

$\dot{x_1} = \dot{x} = x_2$

$y_1 = y$

$y_2 = \dot{y_1} = \dot{y}$

=>
$\dot{x_2} = x_1 + 2 y_2 - 0.9 \frac{x+0.1}{D_1} - 0.1 \frac{x-0.9}{D_2}$

$\dot{y_2} = y_1 - 2 x_2 - 0.9 \frac{y_1}{D_1} - 0.1 \frac{y_1}{D_2}$

I did not realize that there are 4 functions: $y_2(t), y_1(t), x_1(t), x_2(t)$

I still want to solve this by using ode45.

Regards
Rapha

Now we have as state vector:

${\bold{x}}=\left[ \begin{array}{c}x\\y\\ \dot{x}\\ \dot{y} \end{array} \right]$

and derivative:

$\dot{\bold{x}}=\left[ \begin{array}{c} \bold{x}_3\\ \bold{x}_4\\ \bold{x}_1-2\bold{x}_4-0.9(\bold{x}_1+0.1)/D_1-0.1(\bold{x}_1-0.9)/D_2 \\ \bold{x}_2-2\bold{x}_3-0.9\bold{x}_2/D_1-0.1\bold{x}_2/D_2 \end{array} \right]$

where:

$D_1 = \sqrt{(((\bold{x}_1+0.1)^2+\bold{x}_2^2)^3)}$

$D_2 = \sqrt{(((\bold{x}_1-0.9)^2+\bold{x}_2^2)^3)}$

You will need to write a Matlab function to evaluate the derivative and pass that to ode45 (the derivative is now too complicated to be easily passed as an anonymous function).

(You cound simplify the derivative by collecting terms)

CB

The following users thank CaptainBlack for this useful post:
Rapha
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