Today, while studying perturbation method for solving polynomial equations, I seen a problem, here I'm going to write those steps where I have problem $$\epsilon^{1-3p}x_o^3 = \epsilon^{-p}x_o$$ The next step is, which I didn't get, how they reduce above equation to this $$1-3p=-p \ \ \ \ \ \ \ \ \ \ \ \ \to (A)$$ The above problem I face when Author/My Teacher were trying to do singular perturbation to find root of polynomial equation $\epsilon x^3 -x+1=0$ and here they considered a general substitution $x=\frac{x_o}{\epsilon^p}$ for singular perturbation and after putting this substitution in given polynomial equation, they reached to above equation $(A)$ where I'm confused
Now I have another problem which is unsolved and where I'm getting same problem but can't handle because an equation type of above is not there but with an extra term which is $$\epsilon x^4 - x^2+3x-2+\epsilon=0$$ and after using substitution $x=\frac{x_o}{\epsilon^p}$ I get $$x_o^4\epsilon^{1-4p} - x_o^2\epsilon^{-2p}+3x_o\epsilon^{-p}-2+\epsilon=0$$ And in my views dominant terms should be $$x_o^4\epsilon^{1-4p} - x_o^2\epsilon^{-2p}+3x_o\epsilon^{-p}=0 \ \ \ \ \ \to (2)$$ Where I removed smaller values $-2+\epsilon$
Now in above last equation, I'm confused either I must condsider only $2$ dominant term in order to solve above equation for $p$ or there may be more terms and the priority question of mine is that how we can solve above equation $(2)$ for $p$ which is in power?
So, after studying more about my above problem I nearly reached to my answer and now I'm going to share some amount related to my own problem.
In the above question where the problem comes for the solution of equation for the variable $p$ in exponents is Gauge Function exponents solution which I get from these notes Singular Perturbation Expansion for Polynomials See Page No. 5 Undetermined Gauges which is exact my problem. The equation in above problem is $$ϵ^{1−3p}x_o^3=ϵ^{−p}x_o$$ Where we solve for $p$ by equating all possible pairs of Gauge Function exponents. As in this equation we have only one case which is $$1-3p=-p$$ So as next in Question I added another equation having more terms where my question was that how we can solve for $p$ (in exponents) if I have terms more than 2 terms then here is answere
As much terms we have, we will have to solve more cases. As in equation $(2)$ of question $$x_o^4\epsilon^{1-4p} - x_o^2\epsilon^{-2p}+3x_o\epsilon^{-p}=0$$ We have 3 terms so we will solve all possible cases for $p$ P by equating all possible pairs of gauge function exponents and here we will get 3 cases
Case I: $1-4p = -2p$
Case II: $1-4p = -p$
Case III: $-2p = -p$
So in these 3 cases, the case we consider in which singular behavior will removed i.e. where we will have $\epsilon$ in only one term.