# Finding the answer to a differential equation

Given the system of nonlinear differential equations: \begin{cases} \dot{x} = -y-xy \\ \dot{y}= x + x^2, \end{cases} a transformation into polar coordinates of this system can be shown to equal: \begin{cases} \dot{\rho} = 0 \\ \dot{\varphi} = 1 + \rho \cos \varphi .\end{cases} Suppose the initial values are given as:

\begin{align} \begin{pmatrix} x(0) \\ y(0) \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \end{pmatrix}. \end{align} From this I have calculated that $\varphi(0)=\pi$ and $\rho(0)=2$. My question is as follows: how can I find the solution to $\varphi(t)$ knowing that $\rho(t)=2$ (as seen from the fact that the differential equation of $\rho$ is 0 and therefore a constant)? In addition, how can I find $x(t)$ and $y(t)$ after I know what $\varphi(t)$ and $\rho(t)$ are?

• You are now simply solving a linear differential equation for $\phi(t)$ where you can replace $\rho$. This is a limit cycle solution. Apr 11 '17 at 19:36

Hint: Since $\rho=2$, you have $$\frac{d\phi}{1+2\cos\phi}=dt$$ and hence $$\int\frac{d\phi}{1+2\cos\phi}=t+C_1$$ or $$\int\frac{d\phi}{1+2\frac{1-\tan^2\frac{\phi}{2}}{1+\tan^2\frac{\phi}{2}}}=t+C_1$$ By letting $\phi=2\arctan u$, one has $d\phi=\frac{2}{1+u^2}du$ and hence $$\int\frac{2}{1+2\frac{1-u^2}{1+u^2}}\frac1{1+u^2}du=t+C_1$$ or $$\int\frac{2}{3-u^2}du=t+C_1$$ or $$\frac{2}{\sqrt3}\text{arctanh} \frac{u}{\sqrt3}=t+C_1$$ So $$u=\sqrt3\tanh(\frac{\sqrt3}{2}t+C_1)$$ and hence $$\phi=2\arctan\left[\sqrt3\tanh(\frac{\sqrt3}{2}t+C_1)\right]$$
• The devil is in the detail. Apr 11 '17 at 19:43
• @Chinny84, let $\phi=2\arctan u$ and you will get it. Apr 11 '17 at 20:04