$$\ddot y + y + \epsilon y^3 = 0$$ $$y(0) = a$$ $$\dot y = 0 $$
How do I solve this to first order using "ordinary asymptotic power series"? Frankly I don't see that term in the section in my book so I'm confused where to start. If someone could start me off that would hopefully help me see where this should be going.
The key point is to make clear what variable or parameter the power series involves. Is it a power series such as $$\sum_{k=0}^\infty a_kx^k \text{ where }a_k \text{ is function of }\epsilon \text{ ?}$$ or $$\sum_{k=0}^\infty a_k\epsilon^k \text{ where }a_k \text{ is function of }x\text{ ?}$$
I tend to think it's the second, not the first. In this case we assume : $$y(x)=y_0(x)+y_1(x)\epsilon+y_2(x)\epsilon^2+...$$ Put it into the ODE : $$\left(y_0''+y_1''\epsilon+y_2''\epsilon^2+...\right)+\left(y_0''+y_1''\epsilon+y_2''\epsilon^2+...\right)+\epsilon \left(y_0+y_1\epsilon+y_2\epsilon^2+...\right)^3=0$$ This leads to the system of equations : $$\begin{cases} y_0''+y_0=0 \qquad (1)\\ y_1''+y_1+y_0^3=0 \qquad (2)\\ y_2''+y_2+3y_0^2y_1=0 \qquad (3)\\ \text{etc.} \end{cases}$$ Taking account of the boundary condition, once you have solved Eq.(1) for $y_0(x)$, you can put $y_0^3(x)$ into Eq.(2) and solve it for $y_1(x)$. Generally, this what is wanted, no more.
But, if the asymptotic $y(x)\simeq y_0(x)+y_1(x)\epsilon$ isn't a sufficient approximate, one can compute $3y_0^2y_1$ and put it into Eq.(3) , then solve it for $y_2(x)$ . An so on, if necessary.