# Non-linear ODE, plane-polar coordinates.

• spitz
I see what I did wrong. I was supposed to divide by \sin^2\theta+\cos^2\theta, not multiply, of course. Then the last line should read:\dot{\theta}=\frac{r}{-r}=1In summary, to find \dot{r} and \dot{\theta} in this problem, you can use the equations \dot{x}=4x+y-x(x^2+y^2) and \dot{y}=4y-x-y(x^2+y^2) and the relationship between x, y, r, and \theta to calculate \dot{r}=\frac{r}{-r}=1 and \dot{\theta}=\frac
spitz

## Homework Statement

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:
spitz said:

## Homework Statement

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 $\dot{\theta}$ ?

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.

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?

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.

Thank you, sir.

Have I got this second part right?

HallsofIvy said:
Then turn around and multiply the first equation by $sin(\theta)$, the second equation by $cos(\theta)$, then subtract.

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

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

$\dot{\theta}=-1$
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:
$\sin\theta\dot{x}-cos\theta\dot{y}:$

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

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

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

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

Last edited:

## 1. What are non-linear ODEs?

Non-linear ODEs (ordinary differential equations) are differential equations that involve non-linear functions of the dependent variable and its derivatives. In other words, the rate of change of the variable is not directly proportional to the variable itself.

## 2. What are plane-polar coordinates?

Plane-polar coordinates are a coordinate system in two-dimensional space where a point is defined by its distance from the origin (r) and its angle with respect to a fixed reference line (θ). This system is commonly used in polar graphs and for describing circular and spiral motion.

## 3. How are non-linear ODEs solved in plane-polar coordinates?

To solve a non-linear ODE in plane-polar coordinates, we first convert the equation into polar form by substituting the polar coordinates (r and θ) for the Cartesian coordinates (x and y). We then use standard techniques such as separation of variables or substitution to solve the resulting equation.

## 4. What are some applications of non-linear ODEs in plane-polar coordinates?

Non-linear ODEs in plane-polar coordinates have many real-world applications, including describing the motion of planets and satellites in space, modeling biological growth and decay processes, and analyzing the behavior of electrical circuits.

## 5. Are there any limitations to using non-linear ODEs in plane-polar coordinates?

One limitation of using non-linear ODEs in plane-polar coordinates is that they can become more difficult to solve as the equations become more complex. Additionally, some physical systems may not be accurately described by a non-linear ODE, requiring the use of more advanced mathematical models.

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