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.
Best Answer
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.