The following perturbation problem: $y''+ y = \epsilon y^2, y(0,\epsilon)=1, y'(0,\epsilon)=0$
So far I have deduced that: $y_0''+y_0=0, y_0(0)=1, y_0'(0)=0$, and so $y_0(x)= cos(x)$
For the second term: $y_1''+y_1=y_0^2$ with $y_1(0)=0, y_1'(0)=0$
How is the term $y_2(x)=\frac{1}{2}-\frac{1}{3}cos(x)-\frac{1}{6}cos(2x)$ deduced if your applying the method of undetermined coefficient. Unsure of the best particular solution which should be used for the homogenous ODE.