0
$\begingroup$

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.

  • 0
    If the post is about ($*$), then may I suggest that you delete everything that comes before ($*$), or at least edit in something to the effect that everything that precedes ($*$) is completely irrelevant to the question you actually want to ask?2012-10-22

4 Answers 4

1

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.

  • 0
    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.2012-10-21
0

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 $y$

  • 0
    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.2012-10-21
0

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} $.

0

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

TouchingCircles@Origin

Incidentally, it is a special case of bipolar system of coordinates and also geodesics of the hyperbolic plane.