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First Order Nonlinear Ordinary Differential Equation

Chipset3600

Member
Feb 14, 2012
79
Hello, can you guys help me please with this differential equation from Demidovitch book, is to find the solution transforming to polar coordinates :

ode.jpg
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Hello, can you guys help me please with this differential equation from Demidovitch book:

View attachment 1479
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.
 

Chipset3600

Member
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
is to find the solution transforming to polar coordinates.
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.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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.
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
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
I agree.
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Jester mentioned to me, in a PM, that if the author suggested polar coordinates, that it was a typo and the DE is
$$y'= \frac{ \sqrt{x^{2}+y^{2}}-x}{y},$$
in which case switching to polar indeed makes the DE much simpler.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Jester mentioned to me, in a PM, that if the author suggested polar coordinates, that it was a typo and the DE is
$$y'= \frac{ \sqrt{x^{2}+y^{2}}-x}{y},$$
in which case switching to polar indeed makes the DE much simpler.
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 Jester suggests. Before post #3 was edited to mention polar coordinates, I looked at it as possibly meant to be solved using the substitution \(\displaystyle v=\frac{y}{x}\).
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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 Jester suggests. 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 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.

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

-Dan
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
If we start from:
$$y'=\frac{\sqrt{x^2+y^2} - x}{y}$$
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}
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
It's actually even easier than I like Serena's post.

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.