Solving a nonlinear autonomous system using polar coordinates I have the nonlinear system:
\begin{equation}
\begin{array}
fx'=-y-x\sqrt{x^2+y^2}\\
y'=x-y\sqrt{x^2+y^2}
\end{array}
\end{equation}
which I solve by polar coordinates transformation. $x=r\cos\phi \ y=r\sin\phi$ and $x'=dr\cos\phi-r\sin\phi$ and $y'=dr\sin\phi+r\cos\phi$.
\begin{equation}
\begin{array}
fdr\cos\phi-r\sin\phi=-r\sin\phi-r\cos\phi\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}\\
dr\sin\phi+r\cos\phi=r\cos\phi-r\sin\phi\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}\
\end{array}
\end{equation}
Clearly, many terms cancel out and we obtain:
\begin{equation}
\begin{array}
fdr=-r^2\\
dr=-r^2\
\end{array}
\end{equation}
Then, integrating both sides with respect to $r$ we obtain:
\begin{equation}
\begin{array}
fr=-\frac{1}{3}r^3+c\\
r=-\frac{1}{3}r^3+c
\end{array}
\end{equation}
Since $r=\frac{x}{\cos\phi} and r=\frac{y}{\sin\phi}$ we get
\begin{equation}
\begin{array}
fx=-\frac{x^3}{\cos^2\phi}+c\cos\phi\\
y=-\frac{y^3}{\sin^3\phi}+c\sin\phi
\end{array}
\end{equation}
The summing up the two equations, and isolating for y, I get a very scary result:
y≈0.26457 sin^4(ϕ) cos(ϕ) (sqrt((27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^2 + 108 csc^15(ϕ) sec^6(ϕ)) + 27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^(1/3) - (1.2599 csc(ϕ) sec(ϕ))/(sqrt((27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^2 + 108 csc^15(ϕ) sec^6(ϕ)) + 27 c csc^9(ϕ) sec^2(ϕ) + 27 c csc^8(ϕ) sec^3(ϕ) - 27 x^3 csc^9(ϕ) sec^5(ϕ) - 27 x csc^9(ϕ) sec^3(ϕ))^(1/3) and csc(ϕ) sec(ϕ)!=0
What went wrong here?
Thanks
UPDATE, with Jean Maries correction we get:
\begin{equation}
\begin{array}
fx=\frac{\cos\phi}{\phi+c}\\
y=\frac{\sin\phi}{\phi+c}
\end{array}
\end{equation}
Summing up and solving for y:
\begin{equation}
y=\frac{\cos\phi}{\phi+c}+\frac{\sin\phi}{\phi+c}-x
\end{equation}
which plotted , with an arbitrary value of $c$ is:


$dr=-r^2$ cannot be integrated as you do. You need to write it under the form
$$- \dfrac{dr}{r^2}=1$$
giving :
$$\dfrac{1}{r}=\varphi+c$$
($c$ arbitrary constant), otherwise said:
$$r=\frac{1}{\varphi+c}$$
