Why is this function not a solution to this ODE? If we consider the autonomous ODE:
$\frac{dx}{dt} = f(x) = ax$, where $a$ is in $\mathbb{R}$, $x(0) = x_{0}$
then this has a unique solution for every initial condition $x(0) = x_{0}$ since $f$ is continuous and so is $\frac{df}{dx}$, namely:
$x(t) = x_{0}e^{at}$.
But if we just consider the initial condition $x(0) = 0$, fix a constant $b > 0$, and define a function:
$z(t) = e^{a(t - b)}$    if $t > b$,
and   $z(t) = 0$     if $t \leq b$.
Now I differentiated and plugged it into the ODE (not realising there was a discontinuity at $t = b$) and found that it seemed to be a solution to the IVP ($x(0) = 0$) - if this was a solution to the IVP  for any choice of $b$, then it would violate uniqueness of solutions as we would have infinitely many choices of $b$ that we could make - so clearly it is not a solution to the IVP, since we already know $x(t) = 0$ for all $t$ is a solution to the IVP, and $z \neq x$.
But why is $z(t)$ not a solution?
Someone pointed out to me that it was because $z(t)$ was discontinuous at $t = b$ and was clearly not differentiable there - initially I found this satisfactory but then I looked at another example in my lecture notes:
$\frac{dx}{dt} = x^2$, $x(0) = x_{0}$
The solution is $x(t) = \frac{x_{0}}{1 - x_{0}t}$, and it was said that the solution 'blows up' for $t = \frac{1}{x_{0}}$ as we are dividing by 0 there. But this solution is discontinuous and not differentiable at $t = \frac{1}{x_{0}}$ just like $z$, but is being recognised as a solution. Indeed, my lecture notes say the solution is only for 'finite time'. What does this mean?
So do we say $z$ is a solution only for 'finite time'? What would/does this mean, that we ignore the point $t = b$? But then we still have the problem of this being a different solution to $x(t) = 0$.
I am very confused by this and feel I must have misunderstood something somewhere. Hopefully I haven't been too vague or unclear - I'd be happy to clarify anything in the comments! Any help would be greatly appreciated! Thanks!
 A: Your proposed function $z(t)$ is a solution, but only on the interval $t < b$.
Uniqueness means that whenever you have a function $x(t)$ which satisfies the ODE on an open interval $I$ containing the point $t=0$ and satisfies the initial condition $x(0)=0$, then it has to satisfy $x(t)=0$ for every $t$ in that interval $I$.
And this is the case for your function $z(t)$ (for any choice of $b>0$), so it doesn't violate the uniqueness theorem.
(Regarding the finite time stuff, that solution blows up to infinity “automatically”, just by doing what the ODE says that it should do. So it's not just a matter of constructing a function which jumps “artificially”, as in your first example, it's an unavoidable issue which is built into that particular ODE.)
A: $z(t)$ as defined above is a solution for finite time ($t<b$)
But then isn't that the same thing as $z(t) = 0$
A: In ODE theory, a solution is unique not for an ODE, but for an initial value problem (IVP) for an ODE.  The IVP is specified by giving: (1) the ODE, and (2) the initial condition.  Your $z(t)$ fails to satisfy the original initial condition, unless $x_{0}$.
