Exercise: Solve the following nonlinear problem, for $\,0<\epsilon\ll1$ $$ \left\{\, \begin{aligned} &y'' +\frac{y'}{y^2} \epsilon - y' = 0,\qquad -\infty<x<\infty\\[1mm] &y(-\infty) = 1, \ y(\infty)=\epsilon \end{aligned} \right. $$ Then, study the asymptotic behaviour of its solution for $\,x<0\,$ and $\,x\approx 0\,$ using the equation. Check the results comparing with the exact solution.
My solution
To solve the problem, we reduce the order introducing a constant $\,C$ $$ y' - \frac{\epsilon}{y} - y = C. $$ It is easy to see that the general solution does not satisfy the conditions at infinity if we set $\,C=0$. Setting $\,C\neq 0\,$ and knowing $\,y\not\equiv 0$, we can integrate the previous equation as follows $$ \int\frac{y\,dy}{y^2 + Cy + \epsilon} = x + D. $$
We notice that if $\,C^2 - 4\epsilon < 0\,$ then $\,y^2 + Cy + \epsilon > 0\,$ because $$ y^2 + Cy + \epsilon = \left(y + \frac{C}{2}\right)^2 +\epsilon - \frac{C^2}{4} $$ and we can compute the integral as (step by step integration omitted) $$ \int\frac{y\,dy}{y^2 + Cy + \epsilon} = \log\sqrt{y^2+Cy+\epsilon\,} - \frac{C}{\sqrt{4\epsilon - C^2}}\,\text{atan}\!\left(\frac{2y + C}{\sqrt{4\epsilon - C^2}}\right). $$
The expression above does not verify the boundary conditions. Therefore, let's assume the opposite, i.e, $\,C^2-4\epsilon\geq0$.
If $C^2=4\epsilon$, $$ \int\frac{y\,dy}{y^2 + Cy + \epsilon} = \int\frac{y\,dy}{(y+C/2)^2} = -\frac{y}{y+C/2} + \log\,\left|y+\frac{C}{2}\right| $$ but $$ \lim_{x\to-\infty}\exp\left(-\frac{y(x)}{y(x) + C/2}\right)\cdot\left|y(x)+\frac{C}{2}\right| = 0 = \lim_{x\to-\infty}Ke^x $$ if and only if $\,C = -2$. This means $\,\epsilon = 1$, which is a contradiction.
The only possibility left is $\,C^2-4\epsilon>0$, thus $$ y^2 + Cy + \epsilon = (y - y_1)(y - y_2), $$ where $\,y_1 > y_2\,$ and $\,C = - y_1 - y_2$, $\,\epsilon = y_1y_2$.
We have to remark that $\,y(x)>0\,$ for all $\,x\in\mathbb{R}$. Otherwise, it'd exist $x_0$ such that $y(x_0)=0$ and the equation would make no sense. Assuming $\,y_1>y_2>0$, $$ \int\frac{y\,dy}{y^2 + Cy + \epsilon} = \int\frac{A}{y - y_1}\,dy\, + \int\frac{B}{y - y_2}\,dy = \frac{1}{y_1 - y_2}\log\frac{|y - y_1|^{y_1}}{|y - y_2|^{y_2}} $$ and back again, we apply the condition when $\,x\to-\infty\,$ to $$ \lim_{x\to-\infty} \frac{|y(x) - y_1|^{y_1}}{|y(x) - y_2|^{y_2}} = 0 = \lim_{x\to-\infty} Ke^{(y_1 - y_2)x}. $$ We deduce $\,y_1 = 1$, hence $\,y_2 = \epsilon\,$ and $\,C = -1 - \epsilon$. It is easy to prove that the solution verifies the condition at $\,\infty$. Furthermore, considering the roots $y_1, y_2$, the equation has the form $$ yy' = (y-1)(y-\epsilon), $$ and it can be proved $\,y(x)$ stays between $\,y_1$ and $\,y_2$. So, we can assure $$ \frac{1 - y(x)}{(y(x) - \epsilon)^\epsilon} = Ke^{(1-\epsilon)x}. $$
I do not obtain a closed formula for $y(x)$ and the last equality still depends on $K$, so how am I supposed to compare the asymptotic behaviour with the exact solution?
Thanks in advance for the help.