Unable to find Lipschitz constant for $y'=(t-1)\sin(y)$ Given the problem: 
$$y′ = (t − 1)\sin(y),\;\;\;y(1) = 1$$
find an approximation for $y(2)$.
Give an upper bound for the global error taking $n = 4$ (i.e., $h = \frac{1}{4}$)
The goal is to find an upper bound, not actually find an approximation for $y(2)$.
However I don't think it is possible to find an upper bound as we need to use the equation
$$|E_n| \leq \frac{T}{L}[(e^{(t_n - t_0)L} - 1)]$$
where $T$ is the max truncation error and $L$ is the Lipschitz constant.
But I am unable to get a Lipschitz constant from $y′ = (t − 1)\sin(y)$, I don't think it's possible without being given a bound on $t$.
$$|f(t, u) - f(t,v)| = |(t-1)\sin(u) - (t-1)\sin(v)|= |t-1||\sin(u) - \sin(v)|$$
$\sin\;$ has Lipschitz constant of $2$ so we can say
$$\leq 2|t-1||u - v|$$
And where can I go from here...?
Is the question just not well defined or am I missing something?

Letting $f(t,y) = (t-1)\sin(y)$, you can use that an upper bound on $|\partial_y f(t,y)|$ will be a Lipschitz constant (this is an application of the mean value theorem).  To see this, note, as the commenter mentioned, you only need to consider values of $1\leq t \leq 2$.  It should be clear that a bound on this partial exists for any $t$ in this interval, and any value of $y$.
