Question: "Consider the singularly perturbed cubic equation
$\epsilon x^3+x^2-1=0$
where $\epsilon >0$ is a small parameter. Obtain three-term astymptotic expansions for the two regularly perturbed roots as $\epsilon \to 0$. Use the substitution $x(\epsilon) = \frac{z(\epsilon)}{\epsilon}$ to obtain the first two terms of an asymptotic approximation of the singulary perturbed solution of this equation as $\epsilon \to 0$."
Look for a solution of the form
$x(\epsilon) = x_0 + \epsilon x_1 + \epsilon^2 x_2 +...$
formally sub into equation
$\epsilon(x_0 + \epsilon x_1 + \epsilon^2 x_2 +...)^3 + (x_0 + \epsilon x_1 + \epsilon^2 x_2 +...)^2 - 1 = 0$.
Then we have:
$\epsilon(x^3_0+\epsilon(3x^2_0x_1) + \epsilon^2(3x_0x_1^2+3x_0^2x_2)+...)+(x^2_0+\epsilon(2x_0x_1)+\epsilon^2(x_1^2+2x_0x_2)+...)-1=0$.
Comparing coefficients of powers of $\epsilon$:
$O(\epsilon^0) = x^2_0-1=0$
$ \implies x_0=+1,-1$
$O(\epsilon^1) = x^3_0+2x_0x_1=0$
$ \implies $ if $x_0=+1 \space$then $x_1=\frac{-1}{2}$ and if $x_0=-1$ then $x_1 = \frac{1}{2}$
$O(\epsilon^2) = 3x^2_0x_1+x^2_1+2x_0x_2=0$
$ \implies $ if $x_0=+1 \space$ and $x_1=\frac{-1}{2}$ then $x_2=\frac{5}{8}$ and if $x_0=-1$ and $x_1 = \frac{1}{2}$ then $x_2=\frac{-5}{8}$
Giving the final answer of
$x(\epsilon)=1-\frac{1}{2}\epsilon + \frac{5}{8}\epsilon^2+...$
and
$x(\epsilon)=-1+\frac{1}{2}\epsilon - \frac{5}{8}\epsilon^2+...$
This is my solution for the first part, I have then gone onto begin the second part of the question by using the substitution and beginning to solve in the same way.
I have plugged in to give me
$\frac{(z(\epsilon))^3}{\epsilon^2} + \frac{(z(\epsilon))^2}{\epsilon} - 1=0$
then subbing in $x(\epsilon) = z(\epsilon)$ gives
$\epsilon(\frac{z(\epsilon)}{\epsilon})^3$ + $(\frac{z(\epsilon)}{\epsilon})^2 - 1 = 0$
simplifying gives
$\frac{1}{\epsilon^2}(z(\epsilon))^3$ + $(\frac{z(\epsilon)}{\epsilon})^2 - 1 = 0$
then using $z(\epsilon)=z_0+ \epsilon z + \epsilon z_2 +..$ and subbing into the above equation we have
$\frac{1}{\epsilon^2}(z_0^3+\epsilon(3z^2_0x_1)+\epsilon^2(3z_0z^2_1+3z^2_0z_2)+...) + \frac{1}{\epsilon}(z^2_0+\epsilon(2z_0z_1)+\epsilon^2(z_1^2+2z_0z_2)+...) -1 = 0$
Looking at problems at different powers of $\epsilon$:
$O(\epsilon^0) = 3z_0z^2_1+3z^2_0z_2-1=0$.
The above is what I have so far, which I dont believe is correct.
TIA