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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

- Thread starter Chipset3600
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- #1

- Feb 14, 2012

- 79

What are you supposed to do? Find the steady-states? Plot the vector field? Also, in the the notes section, there are notes on Adv ODEs which are about nonlinear ODEs, dynamical systems, and bifurications.Hello, can you guys help me please with this differential equation from Demidovitch book:

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- #3

- Feb 14, 2012

- 79

What are you supposed to do? Find the steady-states? Plot the vector field? Also, in the the notes section, there are notes on Adv ODEs which are about nonlinear ODEs, dynamical systems, and bifurications.

is to find the solution transforming to polar coordinates.

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- #4

- Jan 26, 2012

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Hmm. If it was $x^{2}+y^{2}$ under the square root, I could believe that that method would gain you something. But it's $x^{2}+y^{3}$ - rather asymmetric. I played around with this DE a bit, and couldn't get anything to work. Maybe Jester could weigh in on it.is to find the solution transforming to polar coordinates.

- Aug 30, 2012

- 1,208

I played with various substitutions to get rid of that square root and got nowhere. In frustration I turned to Mathematica (which is usually a bit better than Wolfram|Alpha) and it flatly refused to solve the equation. I tried various versions of the equation and Mathematica didn't give me anything there either. Let the Calculus gurus around here have the final say if it can be solved in closed forms, but I don't think it can be done other than numerically.Hmm. If it was $x^{2}+y^{2}$ under the square root, I could believe that that method would gain you something. But it's $x^{2}+y^{3}$ - rather asymmetric. I played around with this DE a bit, and couldn't get anything to work. Maybe Jester could weigh in on it.

-Dan

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- #6

- Mar 5, 2012

- 9,417

I agree.I played with various substitutions to get rid of that square root and got nowhere. In frustration I turned to Mathematica (which is usually a bit better than Wolfram|Alpha) and it flatly refused to solve the equation. I tried various versions of the equation and Mathematica didn't give me anything there either. Let the Calculus gurus around here have the final say if it can be solved in closed forms, but I don't think it can be done other than numerically.

-Dan

Polar coordinates do not appear to have any use for this DE.

The best we can do is a numerical solution, or analyze some properties of the solution.

Properties I can see is that for $y=0$ and $x\ge0$, the DE is indeterminate.

For $y=0$ and $x<0$ we have a vertical tangent.

And for $y < -|x|^{2/3}$ the solution is undefined.

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- #7

- Jan 26, 2012

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$$y'= \frac{ \sqrt{x^{2}+y^{2}}-x}{y},$$

in which case switching to polar indeed makes the DE much simpler.

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- #8

Like everyone else, I tried various methods with no success, and I too began to wonder if the problems was incorrectly given in the book and should have been given as

$$y'= \frac{ \sqrt{x^{2}+y^{2}}-x}{y},$$

in which case switching to polar indeed makes the DE much simpler.

- Aug 30, 2012

- 1,208

I tried both v = y/x and v = x/y. Both methods gave me something that looked like it might work out to give some kind of elliptic integral but I couldn't finish putting it in that form either. I got a little ways into series expansion but I got a headache before I got too far into it. That square root is a bear for a series solution.Like everyone else, I tried various methods with no success, and I too began to wonder if the problems was incorrectly given in the book and should have been given asJestersuggests. Before post #3 was edited to mention polar coordinates, I looked at it as possibly meant to be solved using the substitution \(\displaystyle v=\frac{y}{x}\).

I'd like to hear what Chipset3600 has to say about the matter at this point.

-Dan

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- #10

- Mar 5, 2012

- 9,417

$$y'=\frac{\sqrt{x^2+y^2} - x}{y}$$

and switch to polar coordinates, then:

\begin{array}{lcl}

dy&=&\frac{\sqrt{x^2+y^2} - x}{y}dx \\

d(r\sin\phi)&=&\frac{r-r\cos\phi}{r\sin\phi}d(r\cos\phi) \\

dr \sin\phi +r\cos\phi d\phi &=& \frac{1-\cos\phi}{\sin\phi} (dr\cos\phi - r\sin\phi d\phi) \\

\cdots \\

\frac{\sin\phi d\phi}{1 - \cos\phi} &=& -\frac {dr}{r} \\

\ln(1 - \cos\phi) &=& -\ln(r) + C_1 \\

1 - \cos\phi &=& \frac {C} r \\

r\cos\phi &=& r - C \\

x &=& \sqrt{x^2+y^2} - C \\

x + C &=& \sqrt{x^2+y^2} \\

x^2 + 2Cx+C^2 &=& x^2+y^2 \\

y^2 &=& 2Cx+C^2

\end{array}

- Jan 26, 2012

- 183

If $x = r \cos \theta$ and $y = r \sin \theta$ then $x^2+y^2 = r^2$, so $x + y \dfrac{dy}{dx} = r \dfrac{dr}{dx}$ and the ODE

$ y \dfrac{dy}{dx} + x = \sqrt{x^2+y^2}$ becomes $r\dfrac{dr}{dx} = r$ and the rest is easy.