Leading behaviour of DE at infinity This is taken from the book of Bender and Orszag, problem 3.44. 
Find the leading behavior as $x\rightarrow+\infty$ of the differential equation:
$x^3y'' - (2x^3 -x^2)y' +(x^3-x^2-1)y=0$
Explain the appearance of a logarithm in the leading behavior. It appears that $e^x$ is a solution and I'm not sure where the logarithm comes from. Usually, one uses the ansatz $y\sim e^{S(x)}$ and does a dominant balance but clearly this method cannot produce a logarithm. 
 A: After the substitution $y=\exp S$, we have the problem
$$ x^3 S'' + x^3 S'^2 +(x^2-2x^3) S' + x^3 -x^2-1=0 \tag{1}$$
Dominant balance is the proper way to approach the problem. For large $x$, (1) reads
$$ x^3(S'' +  S'^2 - 2  S' + 1)=0.\tag{1a}$$  
Now for the dominant balance, we assume that $S'' \ll S'^2$, we obtain the problem
$$ (S'-1)^2 =0\tag{2}$$
with the solution $S= x +\ln c $. This regime is consistent, as $S''=0$. Note however that we obtain only the single solution (as you have observed)
$$S= c e^{x}.$$ So the question is what the 2nd independent solution is.
The problem arises as in (2) $S'=1$ is a degenerate solution. So in order to find a second independent solution, we have to take the perturbation due to $S''$ into account. So, we insert $S= x + \ln c_1 + C(x)$ into (1a) and obtain $$C'^2+C''=0$$ with the solution $C= \ln(x- c_2)$. Thus, we have the asymptotic expression
$$ y \sim \exp[x+\ln c_1 +\ln(x-c_2)] = c_1 e^x (x-c_2).$$
The appearance of the logarithm can be understood by investigating the differential equation
$$y''-2y'+y=0 \tag{3}$$
that is obtained from your differential equation in the limit $x\to \infty$. The linear equation (3) has the general solution
$$ y = c_1 e^{x} +c_2 x e^{x},$$
the logarithm in the expression for $S$ arises due to the prefactor $x$ in the second term (due to the degeneracy of the characteristic values).
