If we model the setup with a capacitor in parallel with the load, and let \$i\$ be the current into the load and \$v\$ the voltage across the two, the circuit equations are:

$$
\left\{
\begin{aligned}
-i &= C \dfrac {dv} {dt}
\\[1 em]
v i &= P 
\end{aligned}
\right.
\qquad \Leftrightarrow \qquad
\left\{
\begin{aligned}
-i &= C \dfrac {dv} {dt}
\\[1 em]
i &= \dfrac P  v
\end{aligned}
\right.
$$

Where P is the constant power level. Putting the two together will give you this differential equation:

$$
C \dfrac {dv} {dt} = - \dfrac P v
\qquad \Leftrightarrow \qquad
v dv = - \dfrac P C dt
\qquad \Leftrightarrow \qquad
2 v dv = -2 \dfrac P C dt
$$

If we integrate starting at instant 0 where we assume a voltage \$v_0\$ is across the cap, we get:

$$
\int_{v_0}^{v} {2 v dv} = \int_0^t {-2 \dfrac P C dt}
\qquad \Leftrightarrow \qquad
v^2 - v_0^2 = -2 \dfrac P C t
$$

From which you can readily get a formula for the voltage and the time:

$$
t = \dfrac {C}{2 P} (v_0^2 - v^2)
\qquad
v = \sqrt{v_0^2 - \dfrac{2P}{C} t}
$$

If we want to take into account the ESR or other circuit elements the equation becomes nastier to solve, but for an initial guess it should be good enough.