# ODE phase portrait

## Homework Statement

i need to create the phase diagram for the equation $$\ddot{x}=x^{3}-x$$
using MATLAB

## Homework Equations

Well i have worked out that there is a centre point at 0,0 and a saddle point at -1,0.

## The Attempt at a Solution

i found a site on the internet which shows me what the diagram should look like:

(http://fds.oup.com/www.oup.com/pdf/13/9780199208241.pdf [Broken]) page 19

but i have no idea how to do this in Matlab.

I tried to do it using a quiver but its not looking anything like this.

Then i tried to do it using ode23 package but i keep getting an error saying "Unable to meet integration tolerances without reducing the step size below the smallest value allowed "

would really really appreciate any help, thanks

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hunt_mat
Homework Helper
To compute the phase portrait, compute:

$$\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=\frac{x^{3}-x}{y}$$

This is an integral which is do-able, so do it and see what you get.

do you mean just straight forward integrate it w,r,t, x
for example it would be

$$\frac{x^4 - 2x^2}{4y}$$ ??

then what would i do with that though??
thanks so much for your help

hunt_mat
Homework Helper
No, the integral wold be:

$$y\frac{dy}{dx}=x^{3}-x\Rightarrow\int ydy=\int x^{3}-xdx$$

i see -- ok sorry.

so
$$\frac{y^2}{2} = \frac{x^4-2x^2}{4} + C$$ ??

i still dont really know what to do with it though forgive me!!

thanks again

hunt_mat
Homework Helper
So now you plot y against x for certain values of C, that is your phase diagram.

for any values of C??
all i am getting is what looks like a square graph translated a lot im not sure what its mean to look like

thanks

hunt_mat
Homework Helper
Look at where there are turning points and if they're max/min.

For different values of C.