Resolution of differential equations by polar coordinates

Question

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.

Answer 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.

Answer 2

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 $y$

Answer 3

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} $.

Answer 4

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.