Poirot said:
solve (first three terms) $y''-(2+x^2)y=0$ using the method of dominant balance (for large x).
I solved this and got 4 different solutions when I think I should be getting 2.
The method here is that you look for a solution of the form $y=e^w$, where $w$ is some function of $x$ that has to be determined. Then $y' = w'e^w$, $y'' = (w'' + w'^2)e^w$, and the equation $y''-(2+x^2)y=0$ becomes $w'' + w'^2 = x^2+2.$
As a first approximation, you try $w' = ax^\alpha$. Then $w'' + w'^2 = a\alpha x^{\alpha-1} + a^2x^{2\alpha}$. We want this to look like $x^2+2$ for large $x$. The dominant term there is clearly $x^2$, and to make this balance with the dominant term in the expression for $w'' + w'^2$, we need to take $\alpha=1$ and $a^2=1$. That gives you two possible answers, one for $a=1$ and one for $a=-1.$
Taking the case $a=1$ (so that $w'=x$), we look for a second term, so that $w'$ is then of the form $w' = x+bx^\beta$ (where $\beta<1$). In that case, $w'' = 1 + b\beta x^{\beta-1}$ and $w'^2 = x^2 +2bx^{\beta+1} + b^2x^{2\beta}.$ Thus $w''+w'^2 = x^2 + 1 +2bx^{\beta+1} + b\beta x^{\beta-1} + b^2x^{2\beta}.$ Comparing this with $x^2+2$, we see that the term $x^2$ already balances. The next thing to look for is the constant term. In the expression for $w''+w'^2$ it is 1, but we want it to be 2. So we need an additional 1, and we can get that by taking $\beta=-1$ and $b=\frac12.$ Thus $w' = x + \frac12x^{-1}.$
In a similar way, for the third term we take $w' = x + \frac12x^{-1} +cx^{\gamma}$ (where $\gamma<-1$). Then $w'' = 1 -\frac12x^{-2} +c\gamma x^{\gamma-1}$ and $w'^2 = x^2 +1 + \frac14x^{-2} + 2cx^{\gamma+1} + cx^{\gamma-1} + c^2x^{2\gamma}.$ Thus $$w''+w'^2 = x^2 + 2 -\tfrac14x^{-2} + 2cx^{\gamma+1} + c(1+\gamma)x^{\gamma-1} + c^2x^{2\gamma}.$$ This time, we already have the $x^2+2$ that we want, but we need to eliminate the unwanted term $-\tfrac14x^{-2}$. To do that, take $\gamma=-3$ and $c=\frac18$, so that the term $2cx^{\gamma+1}$ becomes $\tfrac14x^{-2}$ which balances out the unwanted term.
Thus the three-term approximation for $w'$ is $w' = x+\frac12x^{-1}+\frac18x^{-3}$. (You then need to integrate this to get $w$ and then exponentiate it to get $y$.) Notice that the only stage where there was any choice in the parameters was in the first step, where we took $a=1$ rather than $a=-1$.
Taking $a=-1$ and applying the same method, I get the second solution as $w' = -x -\frac32x^{-1} +\frac{15}8x^{-3}$ (but I haven't checked that and I don't guarantee its accuracy). So there are indeed two solutions, as there should be.