Closed-form solution for capacitor charged with current source and discharged through resistor
I solved it via Laplace (as AJN recommends above).
The circuit in Laplace,

Using superposition to find voltage, V(s), from the upper node to reference:
the contribution from the current source will be,
$$V^ʹ(s)=\frac{I}{s}(\frac{\frac{R}{sC}}{R+\frac{1}{sC}})  $$
the contribution from the capacitor initial voltage (assuming charged to V1 prior to t = 0) will be,
$$V^{ʹʹ}(s)=\frac{V1}{s}(\frac{R}{R+\frac{1}{sC}})  $$
the contribution from the voltage source will be,
$$V^{ʹʹʹ}(s)=\frac{V1}{s}(\frac{\frac{1}{sC}}{R+\frac{1}{sC}})  $$
So,
$$V(s)=V^{ʹ}(s)+V^{ʹʹ}(s)+V^{ʹʹʹ}(s)$$
Inverse Laplace transform,
$$v(t) = \mathcal{L}^{-1}[\frac{I}{s}(\frac{\frac{R}{sC}}{R+\frac{1}{sC}})]+\mathcal{L}^{-1}[\frac{V1}{s}(\frac{R}{R+\frac{1}{sC}})]+\mathcal{L}^{-1}[\frac{V1}{s}(\frac{\frac{1}{sC}}{R+\frac{1}{sC}})]$$
$$v(t) = IR(1-e^{\frac{-t}{RC}})+V1e^{\frac{-t}{RC}}+V1(1-e^{\frac{-t}{RC}}) $$
Which simplifies to,
$$v(t) = IR(1-e^{\frac{-t}{RC}})+V1$$
I plugged in R = 2Ω, C = 100µF (charged to 12V at t = 0), I = 5A, and V1 = 12V and the above solution plots as,