Why is the analytic solution not available?
Have you done the time derivative of $V(x)$?
Hint: Once you compute $\frac{d}{dt}V(x)$ (note that it does not depends explicitly on $t$), you can upper bound it with expressions like $\sin x \leq 1$. And for the worst case condition, find such $u$ in order to make the derivative always negative for some compact set $\mathcal{C}=\{x: V(x) \leq c\}$. Actually, $\mathcal{C}$ of interest is given by the bounds of your state vector in the statement of your problem.
To my knowledge there is not a general method for finding a Lyapunov function. In this case one can solve the differential equations and use that to find a Lyapunov function. Namely $x_2$ is decoupled from $x_1$ and can be shown to have the following solution
$$
x_2(t) = C_1\,e^{-t},
$$
where $C_1$ is a constant and depends on the initial condition of $x_2$. Substituting the above equation into the expression for $\dot{x}_1$ gives
$$
\dot{x}_1 = x_1 (C_1\,e^{-t} -1)
$$
which is a separable differential equation, namely
$$
\frac{dx_1}{x_1} = (C_1\,e^{-t} -1) dt.
$$
Integrating on both sides gives
$$
\log(x_1) = -C_1\,e^{-t} -t+C_2.
$$
Solving for $x_1$ gives
\begin{align}
x_1(t) &= e^{-C_1\,e^{-t} -t+C_2}, \\ &= C_3\,e^{-C_1\,e^{-t} -t}, \\
&= C_3\,e^{-t}\,e^{-C_1\,e^{-t}},
\end{align}
or when using the definition for $x_2$ then it can also be expressed as $x_1(t)=C_3\,e^{-t}\,e^{-x_2}$. So the quantities $x_2$ and $x_1\,e^{x_2}$ will both decay exponentially fast, so the following Lyapunov function can be used
$$
V(x) = x_2^2 + x_1^2\,e^{2\,x_2},
$$
for which it can be shown that its derivative is
$$
\dot{V}(x) = -2\,x_2^2 - 2\,x_1^2\,e^{2\,x_2}.
$$
I will leave proving that $V(x)$ is radially unbounded to you.
Best Answer
I'll take the easy way out and point you to the book where a solution to your problem is given along with its proof: Khalil, Nonlinear Systems.
The short answer to your question is yes. Your Lyapunov function does not have to have an explicit $t$ dependence. The general result goes as follows. If you can find a function $V(t,x)$ that is lower and upper bounded by two positive definite functions $W_1(x)$ and $W_2(x)$ and $\frac{\partial V}{\partial t} + \frac{\partial V}{\partial x}f(x,t) \leq -W_3(x)$ for some positive definite function $W_3(x)$, then the origin is uniformly asymptotically stable. In your case $\frac{\partial V}{\partial t} = 0$ and if your $V(x)$ satisfies the conditions I listed above, then you have uniform asymptotic stability.