Find the asymptotic behavior of solutions $y$ of the equation $$x^5 + x^2y^2=y^6,$$ which tends to $0$ when $x$ tends to $0$.
My solution: if $y=Ax^n$, then $$x^5 + A^2x^{2+2n}=A^6x^{6n}.$$ If $2+2n\ge 5$, we should have $5=6n$, but $2+2\cdot5/6 \not> 5$.
If $2+2n< 5$, we have $2+2n=6n$, $n=1/2$; $A^2=A^6\Longrightarrow A=1$, and $y\sim \sqrt x$ when $x\to 0$.
Am I right? How to find the next terms of the expansion of $y$?
Now lets use a correction term
$$y=\sqrt x + f$$
Insert and use only the first two terms in the binomial expansion
$$x^5+x^2(x+2x^{0.5} f)=x^3+6x^{2.5}f$$
$$x^5=4x^{2.5}f$$
$$f=\frac{x^{2.5}}{4}$$
Now, you may want to insert $y=\sqrt x +\frac{x^{2.5}}{4}+f$
Adding (or multiplying) a correction term and inserting into the equation usually works well.
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Beware, I simply guessed the 0.0275