Flyback output capacitors and extremely high RMS current

Deriving the RMS current for a pulsed ramp signal is easy enough, highly applicable to the problem, and informative.

The capacitor doesn't see the ramp exactly, because zero DC flows through a capacitor by definition. Subtracting out the baseline and repeating the integral, we get:

(spoilers as I strongly encourage doing the homework yourself!)

1. Setup

For diode current, we have the function

$$ i_D(t) = \begin{cases} t \frac{I_\text{pk}}{DT} & 0 \le t < DT \\ 0 & DT \le t < T \\ \end{cases} $$ for peak current \$I_\text{pk}\$, duty cycle \$0 \le D < 1\$, and period \$T\$.

2. Mean Value

We need this, to subtract from diode current, to get the capacitor current: \begin{eqnarray} \hat{I_D} &=& \frac{1}{T}\int_0^T i_D(t) dt = \frac{1}{T}\int_0^{DT} t \frac{I_\text{pk}}{DT} dt = \frac{I_\text{pk}}{DT^2} \int_0^{DT} t\, dt \\ &=& \frac{I_\text{pk}}{DT^2} \left[ \frac{t^2}{2} \right]_0^{DT} = \frac{I_\text{pk}}{DT^2} \left[ \frac{(DT)^2}{2} - \frac{0}{2}\right] \\ &=& \frac{I_\text{pk} D}{2} \end{eqnarray}

3. Root Mean Square

\begin{eqnarray} {I_\text{rms}}^2 T &=& \int_0^{T} i_C(t)^2 dt \\ &=& \int_0^{DT} \left( \frac{t I_\text{pk}}{DT} - \frac{I_\text{pk} D}{2} \right)^2 dt + \int_{DT}^{T} \left( -\frac{I_\text{pk} D}{2} \right)^2 dt \\ \end{eqnarray}

Distribute and integrate term by term; take common constants outside:

\begin{eqnarray} {I_\text{rms}}^2 T &=& {I_\text{pk}}^2 \int_0^{DT} \left( \frac{t^2}{D^2 T^2} - \frac{2t}{2T} + \frac{D^2}{4} \right) dt + \frac{{I_\text{pk}}^2 D^2}{4} \int_{DT}^{T} dt \\ \frac{{I_\text{rms}}^2 T}{{I_\text{pk}}^2} &=& \left[ \frac{t^3}{3 D^2 T^2} - \frac{t^2}{2T} + \frac{t D^2}{4} \right]_0^{DT} + \frac{D^2}{4} \left[ \vphantom{\frac{1}{1}} t \right]_{DT}^{T} \\ &=& \frac{(DT)^3}{3 D^2 T^2} - \frac{(DT)^2}{2T} + \frac{(DT) D^2}{4} - 0 + \frac{D^2 (T - DT)}{4} \\ &=& \frac{DT}{3} - \frac{D^2T}{2} + \frac{D^3 T}{4} + \frac{D^2 T}{4} - \frac{D^3T}{4} \\ &=& DT \left( \frac{1}{3} - \frac{D}{4} \right) \\ I_\text{rms} &=& I_\text{pk} \sqrt{D \left( \frac{1}{3} - \frac{D}{4} \right) } \\ \end{eqnarray}

Since you have \$D\$ = 0.5 and \$I_\text{pk}\$ = 13.73A, \$I_\text{rms}\$ = 4.43A.

As for likely capacitors:

If your ambient operating temperature is low, they can be operated modestly above ratings. The rating arises from self-heating, at maximum internal temperature, giving some specified lifetime. (Notice this isn't the same as maximum ambient temperature, or even case temperature; and they don't usually give R_θCA or temp rise at ratings, so extrapolating ratings isn't the most straightforward, unfortunately! Observationally at least, the temp rise at ratings is modest, say 40°C.)

If you can add slope compensation to the controller, mild CCM will be acceptable, for stable operation.

Consider looking at aluminum polymer capacitors, which offer much higher ripple ratings / lower ESR in much smaller values. A few 100 uF might be sufficient here, and then some additional bulk electrolytic (needing much lower ripple rating) can be used to make up the total (whatever the voltage regulation loop requires).