How to find the period of periodic solutions of the van der Pol equation? The equation $$y''+1.115(y^2-1)y'+y=0$$  has solutions that tend towards periodic solutions and I am asked to enter the period of the periodic solutions. How can I find the period without any boundary conditions? And what is the period?
 A: Answer to the first question "How can I find the period without any boundary conditions? " :
You don't need boundary condition since the limit cycle doesn't depend of them. You can chose any initial conditions.
Answer to the second question " What is the period? " (of the limit cycle) :
$$y''+\mu (y^2-1)y'+y=0$$
For small values of $\mu$ the equation is approximately $y''+y\simeq 0$ which solution is $y=C\:\sin\left(2\pi\frac{t}{T}+\varphi \right)$ where the period $T=2\pi$.
For a first approximate, you can take $\quad T\simeq 2\pi\quad$ until $\mu=1.115$ is not large. See the empirical graph below.
Semi-empirical formula from M.Cartwright :
$$T\simeq \left(3-2\ln(2) \right)\mu+3\frac{2.2338}{\mu^{1/3}} $$
This formula, which isn't convenient for small $\mu$, cannot be used in the present case.
An updated equation is represented on the figure below.
In case of not large $\mu$, expending in series of power of $\mu$ leads to :
$$\frac{dt}{dy}\simeq \frac{1}{\sqrt{4-y^2}}+\frac{\mu}{4}y+\frac{\mu^2}{96}(5y^2-2)\sqrt{4-y^2} +...$$
$$T\simeq 2\int_{-2}^2 \left( \frac{1}{\sqrt{4-y^2}}+\frac{\mu}{4}y+\frac{\mu^2}{96}(5y^2-2)\sqrt{4-y^2} \right)dy$$
$$T\simeq 2\pi\left(1+\frac{\mu^2}{16}\right)\qquad\text{not large }\mu .$$
This analytic approximate formula gives $\quad T\simeq 6.77$
Direct numerical solving of the ODE gives $\quad T\simeq 6.75$
More simply, the rough approximate $\quad T\simeq 2\pi\simeq 6.28\quad$ is not too bad in case of $\mu=1.115$.

