I want to know if $x^*(t) =(\frac{2 - e + e^2}{2 - 2e^2})e^t + (\frac{e - 3e^2}{2 - 2e^2})e^{-t} + \frac{1}{2}te^{-t}$ can contain corner points.
This $x^*(t)$ is the solution to the differential equation $x'' - x = -e^{-t}$, which was the solution to This Problem.
I believe, though I'm not sure, that I must use the Erdmann-Weierstrass corner conditions:
$$(i) f_{\dot{x}}|_{t = t_{c^-}} = f_{\dot{x}}|_{t = t_{c^+}}$$
$$(ii) [f - \dot{x}f_{\dot{x}}]_{t = t_{c^-}} = [f - \dot{x}f_{\dot{x}}]_{t = t_{c^+}}$$
For $(i)$, using my value for $f_{\dot{x}}$ that I obtained from the link I provided, I got the following:
$$2(\dot{x}(t) + 1)|_{t = t_{c^-}} = 2(\dot{x}(t) + 1)|_{t = t_{c^+}}$$
$$\dot{x}(t_{c^-}) = \dot{x}(t_{c^+})$$
$$\dot{x}(t_{c^-}) - \dot{x}(t_{c^+}) = 0$$
At this point, I'm thinking that there cannot be any corner points, but I am a bit confused about how to apply the corner conditions and whether those corner conditions are even what I need to show if $\dot{x}(t)$ can have corner points.
$x^*(t)$ doesn't contain any corner, since $x^*(t)$ is continuously differentiable, because each term of it has this property.