Unique time reachability for two symmetric ODE We consider the following ODE $y'(t)=f(t,y(t))$ with initial condition $y(0)=0$ and $y'(t)=-f(t,y(t))$  with initial condition $y(0)=1$ and denote the solutions by $y^+$ and $y^-$ respectively. 
Assume that f is strictly positive and the solution always exists. 
Let $T^-$ and $T^+$ be the unique positive constants such that $y^-(T^-)=0$ and $y^+(T^+)=1$.
My question is: Is $T^-=T^+$? .
Thank you.
 A: I'll assume that $f$ is nice enough for the solution of the initial value problem to be unique, otherwise $y^+$ and $y^-$ aren't well-defined anyway.
It's true that $T^+=T^-$ if the system is autonomous, i.e., if $f(y)$ doesn't depend on $t$ explicitly. Indeed, given $y^+$ (and hence $T^+$), let $\tilde y(t) = y^+(T^+-t)$. Then $\tilde y$ satisfies the same ODE and initial condition as $y^-$, so by uniqueness they are equal:
$$
y^-(t) = \tilde y(t) = y^+(T^+-t)
.
$$
Hence $y^-(T^+)=y^+(0)=0$, so that $T^-=T^+$.
But it's false in general. Just consider the following scenario: For small $t>0$ the function $f(t,y)$ is very large near $y=0$ and almost zero near $y=1$. Then $y^+$ will increase quickly from zero to begin with, while $y^-$ will stay almost constant at $1$. Then, after the solution curves for $y^+$ and $y^-$ cross (say at some value $y=C$, which is just a little less than $1$) the function $f(t,y)$ becomes large for $y>C$ and small for $y<C$. Then $y^+$ will continue to increase quickly, and hit $1$ (at $t=T^+$) long before the slowly decreasing $y^-$ has made it down to zero.
