<s>[INST] Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$ I'm trying to compute the asymptotic expansion for each of the four roots to the following equation, as $\epsilon \rightarrow 0$:
$\epsilon^2x^4-\epsilon x^3-2x^2+2=0$
I'd like my expansions to go up through terms of size $O(\epsilon^2)$.
I´ve made the change of variables $x=\delta y$, performed dominant balance and found out that the only two valid options are: (rescaled eq: $\epsilon^2\delta^4y^4-\epsilon \delta^3 y^3-2\delta^2 y^2+2=0$)


*

*$\delta^2 \sim 2 \Rightarrow \delta=O(1)$

*$\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta=O(\epsilon^{-1})$


How do I proceed?
 [/INST] Two of the roots will be near the roots $1$ and $-1$ of $-2x^2 + 2$, the other two will be near $\infty$.  Solving numerically for $\epsilon = .01$,
those roots are approximately $-100$ and $200$.
So it looks like you want $x \sim c \epsilon^{-1}$.  Indeed with $x = y/\epsilon$ the equation becomes $y^4 - y^3 - 2 y^2 + 2 \epsilon^2 = 0$.
EDIT: At $\epsilon = 0$ the second equation becomes $y^4 - y^3 - 2 y^2 = 0$, which has roots $y = -1$ and $y = 2$ as well as $y=0$ (which corresponds to $x=1$ and $x=-1$).  For the solution near $y=-1$, for example, take $y = -1 + v$ and expand to get $ v^4 - 5 v^3 + 7 v^2 - 3 v + 2 \epsilon^2 = 0$ where $v$ should be small.  To first order, $v = (2/3) \epsilon^2$.  Take $v = (2/3) \epsilon^2 + w$
and the equation becomes 
$$
{w}^{4}+ \left( \dfrac{8 \epsilon^2}{3}-5 \right) {w}^{3}+ \left(\dfrac{8 {
\epsilon}^{4}}{3}-10\,{\epsilon}^{2}+7 \right) {w}^{2}+ \left( {\frac {32
\,{\epsilon}^{6}}{27}}-{\frac {20\,{\epsilon}^{4}}{3}}+{\frac {28\,{
\epsilon}^{2}}{3}}-3 \right) w+{\frac {16\,{\epsilon}^{8}}{81}}-{
\frac {40\,{\epsilon}^{6}}{27}}+{\frac {28\,{\epsilon}^{4}}{9}} = 0
$$
from which $w = (28/27) \epsilon^4 + \ldots$.  Thus this solution is
$$ x = -\epsilon^{-1} + \dfrac{2}{3} \epsilon + \dfrac{28}{27} \epsilon^3 + \ldots$$
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