# how to use regular perturbation method to solve this nonlinear ode?

I want to use the perturbation method to find a solution to this nonlinear ODE:

$$y'' + \epsilon y' + sin(y)=0$$

I have this so far:

attempt $$u(t) = u_0 + \epsilon u_1 + \epsilon^2 u_2 + ...$$ as a solution.

After substituting, using Taylor expansion for $$sin(u)$$, and sorting, I have the following:

$$\epsilon^0$$ term: $$u_0'' + sin(u_0)=0$$.

I can see 1 solution to this ODE which is $$u_0(t)=sin(t)+C$$. Is there a systematic way to solve this kind of equation?

$$\epsilon^1$$ term: $$u_1'' +u_1 + u_0' - \frac{1}{6} \pmatrix{3\\2} u_0^2 u_1 + ... =0$$.

I am not sure if this is correct. Please help.

• Your solution to the zeroth order ODE is incorrect. The technique you might be looking for here is called "two timing." Apr 22, 2022 at 5:03

Your zeroth order equation can be solved by multiplying by $$y'$$ to get: $$1/2dy'^2/dt=d/dt(\cos(y(t))$$, by integrating you get: $$1/2y'^2=\cos(y(t))+E$$ and then integrating once more to get:
$$y=\int \sqrt{2\cos(y(t))+2E}dt$$. Now obviously you can plug instead of $$\cos(y(t))$$ its Taylor series, and then expand with a Taylor series of $$\sqrt{1+x}$$...