Resolution of differential equations by polar coordinates

I have a doubt respect to resolution of differential equations, for example if we have the family of circles $x^2+y^2=2cx$, deriving $$2x+2y\frac{dy}{dx}=2c$$, combining $$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$ and replacing $\frac{dy}{dx}=-\frac{dx}{dy}$, then we have the differential equation $$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}(*)$$ We cannot find the solution of the last differential equation by separation of variables, but if we use polar coordinates in $x^2+y^2=2cx$ we get $$(r\cos\theta)^2+(r\sin\theta)^2=2c(r\cos\theta)$$ and $r=2c\cos\theta$, then $$\frac{dr}{d\theta}=-2c\sin\theta$$ and then $$\frac{r d\theta}{dr}=-\frac{\cos\theta}{\sin\theta}$$

The solution of the last differential equation is $r=2c\sin\theta$, so the solution of the differential equation (*) is $x^2+y^2=2cy$. Then my question is: why the change of coordinates permit find the solution of the differential equation? This is an accident or exit a theorem about this? And if exits such theorem, what kind of differential equation can be solve by change of coordinates? Thanks.

• Somehow, you have gone from $x^2+y^2=2cx$ to $x^2+y^2=2cy$. – Gerry Myerson Oct 21 '12 at 5:02
• I'll add more information. – José Ramírez Oct 21 '12 at 5:07
• You have added more information, but you have not resolved the contradiction. In the second line, you have $x^2+y^2=2cx$. Later, you have $x^2+y^2=2cy$. These are different. One of them has an $x$ where the other has a $y$. How can both of them be the solution, when they are manifestly not equal to each other? – Gerry Myerson Oct 21 '12 at 5:23
• @GerryMyerson: I think the problem is the step in which $$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\mathrm{d}x}{\mathrm{d}y}$$ – robjohn Oct 21 '12 at 10:43
• @robjohn, yes, that step looks highly problematical, especially since what has actually been done looks more like $dy/dx=-1/(dy/dx)$. – Gerry Myerson Oct 21 '12 at 11:48

Solving (*)

We can solve $(\ast)$ without changing to polar coordinates $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2xy}{x^2-y^2}\tag{1}$$ which can be manipulated to $$\frac{\mathrm{d}}{\mathrm{d}y}\frac{x^2}{y}=\frac{2x}{y}\frac{\mathrm{d}x}{\mathrm{d}y}-\frac{x^2}{y^2}=-1\tag{2}$$ Integrating $(2)$ yields $$\frac{x^2}{y}=2c-y\tag{3}$$ For some $c$. Therefore, $$x^2+y^2=2cy\tag{4}$$

Answer to the Question

A change of coordinate may make the solution more apparent, but the equation should be solvable in either coordinate system.

• The equation $$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}(*)$$ is ok, but you don't answer the main question. – José Ramírez Oct 21 '12 at 19:52
• The main question was: "why [does] the change of coordinates permit [us to] find the solution of the differential equation?" My answer was "A change of coordinate may make the solution more apparent, but the equation should be solvable in either coordinate system." – robjohn Oct 21 '12 at 21:25
• The original equation $x^2+y^2=2cx$ has nothing to do with $(\ast)$ as written. That is the point which was confusing to me, and possibly to Gerry Myerson. – robjohn Oct 21 '12 at 21:28

You can use a substitution for the Bernoulli DE you obtain when considering $\frac{dx}{dy} = \frac{x^2 - y^2}{2xy}$

Edit: Notice $\frac{dx}{dy} = \frac{x^2 - y^2}{2xy} \Rightarrow x' - \frac{x}{2y} + \frac{y}{2x} = 0$, where $x'$ is obviously with respect to $y$

• Could you be more expecific? That is not the exact form like here en.wikipedia.org/wiki/Bernoulli_differential_equation – José Ramírez Oct 21 '12 at 4:50
• It appears to be for $n=-1$ where $$P(y)=\frac1{2y}\qquad\text{and}\qquad Q(y)=\frac x2$$ – robjohn Oct 21 '12 at 10:54
• But i know the answer of this, the problem is about this equation $$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}(*)$$ and change of coordinates, you don't answer the main question. – José Ramírez Oct 21 '12 at 19:56

When you replace $\frac{dy}{dx}=-\frac{dx}{dy}$, you get

$$\frac{dx}{dy}=-\frac{y^2-x^2}{2xy}= -\frac{y}{2x}+\frac{x}{2y} \rightarrow (1)\,.$$

It is easier to solve the differential equation (1). Let

$$u=\frac{x}{y} \implies x=yu \implies \frac{dx}{dy}=u+y\frac{du}{dy} \,.$$

Substituting back in $(1)$ gives,

$$u+y\frac{du}{dy}= -\frac{1}{2u}+\frac{u}{2}\implies y\frac{du}{dy} =\frac{3u^2-1}{2u}\,.$$

Now, I think you can solve the last ode by the method of separation of variables to find $u$ as a function in $y$, then, substitute back $u=\frac{x}{y}$.

I saw this only now ... appears not to have been answered to your satisfaction.

You are not at all changing coordinates!! You are finding orthogonal trajectories of

$$x^2 + y^2 + 2 c_1 x =0$$ as

$$x^2 + y^2 + 2 c_2 y =0.$$

with transformation $y^{'}$ to $-1/y^{'}$ on the same coordinate axes, which is the standard procedure to find orthogonal trajectories by a changed differential equation on same coordinate axes of same graph sheet (Red changes to Blue).

$c_1, c_2$ are parameters in each set of circles, displacement on axis = radius of circle. I shall not repeat the integration steps.

Their polar equations are

$$r = 2 c_1 \cos\theta,\;\; r = 2 c_1 \sin \theta$$

The fist one is a set of circles centered on x-axis touching y-axis and passing through origin and the second is the orthogonal set of circles centered on y-axis, also passing through the origin and touching x-axis as shown in the sketch Incidentally, it is a special case of bipolar system of coordinates and also geodesics of the hyperbolic plane.