# Solving $y''+\dfrac{\varepsilon y'}{y^2} - y' = 0, \, y(-\infty)=1$ and $y(\infty) = \varepsilon$

$y''+\dfrac{\varepsilon y'}{y^2} - y' = 0, \, y(-\infty)=1$ and $y(\infty) = \varepsilon >0$.

Above is an ODE that was in an asymptotic analysis exam. A part of the question asked us to solve it exactly on the domain $(-\infty, \infty)$, where I got into troubles. I wrote the equation as: $$\left(\dfrac{\varepsilon}{y}\right)' = y''-y'=(y'-y)'$$ So we have $\dfrac{\varepsilon}{y} - y'+y = C_0$ and then $\dfrac{ydy}{y^2-yC_0+\varepsilon} = dx$, which would give a closed formula for the general solution. But this is where the trouble is: the antiderivative of the LHS will be very different functions depending on the relationships between $\varepsilon$ and $C_0$; therefore, it would seem like we need to determine $C_0$ to begin with. But that is impossible because of the initial conditions not involving any $y'.$

How should I proceed from here?

## 1 Answer

The key hypothesis here (slightly hidden, but still) is that $y(x)$ must be defined for every $x$ in the real line, with prescribed limits when $x\to\pm\infty$.

Thus, the denominator $z^2-C_0z+\epsilon$ must be nonzero for every $z$ between $y(-\infty)$ and $y(+\infty)$ and go to zero at the boundaries $z\to y(-\infty)$ and $z\to y(+\infty)$.

One is given $y(-\infty)=1$ hence $C_0=1+\varepsilon$. Thus, $z^2-C_0z+\varepsilon=(z-1)(z-\varepsilon)$ and $$(1-\varepsilon)x=\int_{y(0)}^{y(x)}\frac{(1-\varepsilon)z}{z^2-C_0z+\varepsilon}dz=\int_{y(0)}^{y(x)}\left(\frac1{z-1}-\frac1{z-\varepsilon}\right)dz$$ For every $y(0)$ between $\varepsilon$ and $1$, this is solved by $$e^{(1-\varepsilon) x}\frac{y(0)-1}{y(0)-\varepsilon}=\frac{y(x)-1}{y(x)-\varepsilon}$$ that is, $$y(x)=\frac{\varepsilon(y(0)-1)e^{(1-\varepsilon) x}-(y(0)-\varepsilon)}{(y(0)-1)e^{(1-\varepsilon) x}-(y(0)-\varepsilon)}$$ For example, if $\varepsilon=\frac12$, one gets $$y(x)=\frac{(1-y(0))e^{x/2}+2y(0)-1}{2(1-y(0))e^{x/2}+2y(0)-1}$$ for every $y(0)$ in $(\frac12,1)$.