# Homework Help: Non-linear ODE, plane-polar coordinates.

1. Mar 13, 2012

### spitz

1. The problem statement, all variables and given/known data

I have:

$\dot{x}=4x+y-x(x^2+y^2)$
$\dot{y}=4y-x-y(x^2+y^2)$

And I need to find $\dot{r}$ and $\dot{\theta}$

2. The attempt at a solution

I got as far as:

$\dot{x}=r(\text{sin}(\theta)-\text{cos}(\theta)(r^2-4))$
$\dot{y}=r(-\text{sin}(\theta)(r^2-4)-\text{cos}(\theta))$

How do I go from here to $\dot{r}$ and $\dot{\theta}$ ?

Last edited: Mar 13, 2012
2. Mar 13, 2012

### LCKurtz

For the left sides of these last two equations you have $x = r\cos\theta,\ y = r\sin\theta$ to work with. Remembering that everything is a function of $t$, you can caculate $\dot x$ and $\dot y$ by differentiating those, using the chain rule.

3. Mar 13, 2012

### spitz

Ok, so I have:

$\dot{r}\cos(\theta)-r\sin(\theta)\, \dot{\theta}=r(\sin(\theta)-\cos(\theta)(r^2-4))$

$\dot{r}\sin(\theta)+r\cos(\theta)\dot\theta=r(-\sin(\theta)(r^2-4)-\cos(\theta))$

Is there a "non-tedious" way of solving this?

4. Mar 13, 2012

### HallsofIvy

I can't speak for what you would consider tedious but the standard method would be to multiply the first equation by $cos(\theta)$, the second equation by $sin(\theta)$, and add the two equations. Then turn around and multiply the first equation by $sin(\theta)$, the second equation by $cos(\theta)$, then subtract.

5. Mar 13, 2012

### spitz

Thank you, sir.

6. Mar 13, 2012

### spitz

Have I got this second part right?

You mean, I presume, $\dot{r}$, not r, on the left of both of those two equations. Now, how did you get "-1" on the right of the last equation? It looks to me like you have r on the right of both first two equations. Isn't r- r= 0?

Last edited by a moderator: Mar 13, 2012
7. Mar 13, 2012

### spitz

$\sin\theta\dot{x}-cos\theta\dot{y}:$

$\dot{r}\sin\theta\cos\theta-r\sin^2\theta\dot{\theta}-\dot{r}\sin\theta\cos\theta-r\cos^2\theta\dot{\theta}=r[\sin^2\theta+\cos^2\theta+(sin\theta\cos\theta-\sin\theta\cos\theta)(4-r^3)]$

$-r(\sin^2 \theta+\cos^2 \theta)\dot{\theta}=r(\sin^2 \theta+\cos^2 \theta)$

$-r\dot{\theta}=r$

$\dot{\theta}=\frac{r}{-r}=-1$

Last edited: Mar 13, 2012