I am unsure of how to interpret the question:
Given $y''+(1+\epsilon)y=0, y(0)=0,y'(0)=1$. Solve exactly.
The context of the problem is that we are practising solving IVP with regular perturbation and Poincare-Lindstedt perturbation.
So would it be correct to assume that I would need ignore $\epsilon$ term, and then to solve it exactly just as:
$y''+y=0, y(0)=0,y'(0)=1$
Which would get $y=sin(x)$. Graphing my solution with Poincare along side it shows "similar" plots so I am guessing that would be correct?
If $a>0$ then the general solution of $y''+ay=0$ is $c_1\cos(\sqrt a x)+c_2\sin (\sqrt a x)$. Here $a=1+\epsilon$. Now can you use the initial conditions to get the solution?
The answer is $\frac {\sin(\sqrt{1+\epsilon}x)} {\sqrt {1+\epsilon}}$.