这是一份southampton南安普敦大学MATH1054W1-01作业代写的成功案例
As in the harmonic balance method, we neglect the effect of the higher harmonic, and see that since $-a \omega \sin \omega t=\dot{x}$ on the limit cycle, can be written as
$$
\varepsilon\left(x^{2}-1\right) \dot{x} \approx \varepsilon\left(\frac{1}{4} a^{2}-1\right) \dot{x}
$$
We now replace this in the differential equation to give the linear equation
$$
\ddot{x}+\varepsilon\left(\frac{1}{4} a^{2}-1\right) \dot{x}+x=0 .
$$
The non-periodic solutions are spirals in the phase plane. Consider the motion for which $x(0)=a_{0}, \dot{x}(0)=0$ : for the next few ‘cycles’ $a_{0}$ will serve as the amplitude used in the above approximation, so may be written
$$
\ddot{x}+\varepsilon\left(\frac{1}{4} a_{0}^{2}-1\right) \dot{x}+x-0 .
$$
With the initial conditions given, the solution is
$$
x(t)=\frac{a_{0}}{\beta} \mathrm{e}^{\alpha t}[\beta \cos \beta t-\alpha \sin \beta t]
$$
where
$$
\left.\alpha=\frac{1}{2} \varepsilon\left(1-\frac{1}{4} a_{0}^{2}\right), \quad \beta=\frac{1}{2} \sqrt{4}-\varepsilon^{2}\left(1-\frac{1}{4} a_{0}^{2}\right)\right]
$$
MATH1054W1-01 COURSE NOTES ：
Therefore $\operatorname{sgn}(x)$ is replaced by $(4 / \pi) \cos \omega t$, which in turn is replaced by $(4 / \pi) x / a$. The equivalent linear equation is then
$$
\ddot{x}+\frac{4}{\pi a} x=0 .
$$
The solution, having any amplitude $a$, of the form $a \cos \omega t$ is
$$
x(t)=a \cos \left[\left(\frac{4}{\pi a}\right)^{1 / 2} t\right]
$$
Therefore
$$
\omega=\frac{2}{\sqrt{(\pi a)}}
$$