Suppose we have the separable ODE,
$$ \frac{dy}{dt} = e^t\frac{y^2-9}{2y}, \;\; y(0) = -5 $$
The explicit solution to this problem is,
$$ y = -\sqrt{9 + 16e^{e^t-1}} $$
I'm confused about the behavior of the solution when $t \rightarrow -\infty$. According to the explicit solution, $y \rightarrow -\sqrt{9 + 16e^{-1}}$. However, if we do a phase line analysis, we can show that $y = -3$ is an unstable stationary point. So as we go back in time, shouldn't, $y \rightarrow -3$? Why do the answers not match?
Does phase line analysis not work for non-autonomous equations like above?
It doesn't work for nonautonomous systems. Consider the change of variables $x = e^t$. Then by the chain rule:
$$\frac{dy}{dt} = \frac{dx}{dt}\frac{dy}{dx} = e^t \frac{dy}{dx}$$
Then the equation reduces down to
$$\frac{dy}{dx} = \frac{y^2-9}{2y}$$
and notice that the final solution becomes
$$y = -\sqrt{9+16e^{x-1}}$$
which does approach the unstable fixed point just like you expected. One to think of our change of variables is that we parametrized the path of the flow lines, but in a way that doesn't retrieve the whole line.