Approaching this differential system using pertubation theory I am not really good at pertubation theory but I found this off some notes and It looks simple but I cannot seem to solve it.
Given the system
\begin{align}
\begin{cases}
x'(t) = y(t) - \epsilon \sin x(t)\\
y'(t) = x(t)y(t) + \epsilon y^2 (t)
\end{cases}, \ x(0)=1, \ y(0)=0, \ 0 < \epsilon << 1
\end{align}
we are asked to find an $\mathcal{O}(\epsilon)$ solution. (Hint: You can use that $\sin (x) = \sum_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n+1}}{(2n+1)!}$.)
Now, this looks simple to my eyes since all I must do is group same things in term of $\epsilon$ and zero in on them. Let's do this
\begin{align}
&x'(t) - y(t) = 0 \land \sin x(t) = 0\\
&y'(t) - x(t)y(t) = 0 \land -y^{2}(t)=0
\end{align}
but I must also start from the inside out. So first line becomes
\begin{align}
x'(t) = y(t) \implies x(t) = \int y(t)dt + C\\
\sin x(t) = 0 \implies x(t) = n\pi \implies y(t) =0
\end{align}
same with the other one, it cancels immediately and rolls down to the trivial solutions. I know this is wrong.
Any help?
 A: It doesn’t work like that – you need to expand the solution in $\epsilon$ and substitute it as a whole:
$$
x(t)=\sum_kx_k(t)\epsilon^k\;,\\
y(t)=\sum_ky_k(t)\epsilon^k\;,
$$
and then
\begin{eqnarray}
\sum_kx_k'(t)\epsilon^k&=&\sum_ky_k(t)\epsilon^k-\epsilon\sin\sum_kx_k(t)\epsilon^k\;,\\
\sum_ky_k'(t)\epsilon^k&=&\sum_kx_k(t)\epsilon^k\sum_ky_k(t)\epsilon^k+\epsilon\,\left(\sum_ky_k(t)\epsilon^k\right)^2\;.
\end{eqnarray}
Now you can read off the equations for each order separately and solve them in turn. For $\epsilon^0$ (from now on dropping the time dependence in the notation):
\begin{eqnarray}
x_0'&=&y_0\;,\\
y_0'&=&x_0y_0\;.
\end{eqnarray}
Differentiating the first equation and substituting into the second yields
$$
x_0''=x_0x_0'\;,
$$
integrating that yields
$$
x_0'=\frac12x_0^2+C\;,
$$
and dividing by the right-hand side and integrating again yields
$$
\int\frac{x_0'}{\frac12x_0^2+C}\mathrm dt=\int1\mathrm dt\;,\\
\sqrt\frac2C\arctan\frac{x_0}{\sqrt{2C}}=t-t_0\;,\\
x_0=\sqrt{2C}\tan\left(\sqrt\frac C2\left(t-t_0\right)\right)\;,
$$
so
$$
y_0=x_0'=\frac C{\cos^2\left(\sqrt\frac C2\left(t-t_0\right)\right)}\;.
$$
(I won’t bother to use the initial values to determine the constants; they don’t seem to lead to anything nice.)
For $\epsilon^1$:
\begin{eqnarray}
x_1'&=&y_1-\sin x_0\;,\\
y_1'&=&x_0y_1+y_0x_1+y_0^2\;.
\end{eqnarray}
Solving the first equation for $y_1$, differentiating and substituting into the second yields
$$
x_1''+x_0'\cos x_0=x_0(x_1'+\sin x_0)+y_0x_1+y_0^2\;,
$$
a second-order ODE for $x_1$ into which you can substitute the known functions $x_0$ and $y_0$. That looks rather intractable; I’ll let you take it from there.
A: Start like the answer of @joriki, but take the initial conditions into account from the start.
The IC $y_0(0)=0$ in the basis stage gives that the IVP starts in a stationary point, giving the constant solution $(x_0(t),y_0(t))=(1,0)$.
This inserted into the first correction gives the system
\begin{align}
x_1'(t)&=y_1(t)- \sin(1),& x_1(0)&=0,\\
y_1'(x)&=y_1(t),&y_1(0)&=0,
\end{align}
resulting in first $y_1(t)=0$ and then $x_1(t)=-\sin(1)t$.
For the second correction the system is
\begin{align}
x_2'(t)&=y_2(t)-\cos(1)\sin(1)t,&x_2(0)&=0,\\
y_2'(t)&=y_2(t),&y_2(0)&=0.
\end{align}
Again immediately $y_2(t)=0$ follows, giving then $x_2(t)=-\frac14\sin(2)t^2$.
One could come to the hypothesis that $y_k(t)=0$ always, thus $y(t)=0$. Then $x(t)=f(ϵt)$ where $f$ is the solution to the IVP $f'(s)=-\sin(f(s))$, $f(0)=1$.
