Mark mentions that DC bias does affect losses and goes on to mention other loss factors which are taken into account when the core loss with frequency and flux excursion are taken into account. So much of it is already covered apart from the DC bias... but for practical purposes you get to ignore that. If you are operating a core, with small ripple, so close to saturation then your design will likely have other problems.

It's magnetics design. Get close to a wrong answer first and then iterate through your design choices.

You're thinking, trying or looking to minimise inductor ripple current, is kind of the wrong way around. In bulk terms Inductors are rubbish at storing energy when compared to capacitors. Designing a large inductor with small ripple is asking for high energy storage. Large inductors, any inductor but in particular large ones, also don't like having the current through them changed too quickly. It really really hurts the speed of your regulation loop if you implement feedback and hurts the output response without feedback.

The 20-40% ripple current is one of those general close to ideal compromises between losses and transient performance, and cost. You don't design output filters in an SMPS based on the concepts of small signal filters you target output ripple voltage and loop performance. Don't get too precious and worry over much about details. Find a result first then check and iterate to the solution.

Your goal is not 5% ripple current. The inductor will be too big and the loop horribly slow as a result. It will spend most of it's time against the stops as it tries to slew the inductor current in response to transient demands. 20% ripple will be four times as fast. 40% ripple eight times as fast. Fast is good.

You goal is output ripple voltage and output voltage deviation in response to transient load demands. That is determined in the first instance by output filter capacitance ESR and in the second by... output filter capacitance ESR. Don't chase esoteric low/zero ESR small value film capacitors. The loop becomes a pain.

Just pick your easiest to source range of radial low ESR 105C aluminium electrolytics with meaningful Data Sheets. What used to be Philips, then BHC and now Vishay 136 RVI. Random...

Low ESR | Aluminum Electrolytic | Capacitors | Vishay
https://www.vishay.com/docs/28321/136rvi.pdf
Your output current is 10A. Your ripple current, 20%, is 2A. You want, can accept, 50mV output ripple voltage. You want an ESR of 50mV/2A or 25mR. Your input voltage is 13V. Assume that if things go wrong your capacitor is going to see that voltage. Pick a voltage rating above this, 16V or to be really safe 25V. Find a capacitor that matches the required ESR and has a 2A ripple current rating. 25V 2200u 16X31. Use two for the extra feel good factor. Make sure you lay out the board so they see the same. Anything tight and thick enough. Check the graphs on the data sheet for frequency/temperature dependencies. This is why you chose a range of capacitors that came with a good, comprehensive, data sheet.

Don't use formulae that include the number 4.44 Those are for sinusoidal excitation voltages. If someone gives you a formula with 4.44 in it. Ignore them. You are dealing with square wave voltages.

Assume Bsat is 300mT. With 20% ripple current your peak to peak flux excursion is 20% of this or 60mT. Half that for the peak and the number you plug in the Pv graph is 30mT. For any, ferrite, material worth its salt you will get a Pv value down in the hundreds, tens or less kW/m^3. The core loss can effectively be ignored. With convection cooling in a transformer design you might start to worry if you are reaching 1 thousand kW/m^3. Notice how no formula was involved.

Your target ripple current is 2A. During switch off time the inductor gets reset through the output voltage plus a diode drop or 7.6V. Your duty cycle is about 50% so the off time is about 3.33uS. Yes you could be more accurate with the sums but ¯\_(ツ)_/¯

dI/dT = Voff/L

L = Voff.Toff/dI

L = 7.6x3.33E-6/2

L = 12.7uH

Fudge equation. These never work. Sometimes they do but many times the estimate turns out to be silly and you have to make other choices.

AwAe = LIpkIrms/BsatJKcu

AwAe is an area product. Winding Area/Core Area. L is your inductor value. Ipk is the average output current plus half ripple, 10A + 1A = 11. Irms is the rms inductor current. Your average or 10A. Bsat = 300mT. J is the wire current density. 4E6 A/m^2. Kcu is copper utilisation. 0.7 for round wire.

AwAe = 13uH x 11 x 10/0.3 x 4E6 x 0.7

AwAe = 1.702E-9

Assume Aw = Ae = SQRT 1.702E-9 = 42 mm^2

https://www.tdk-electronics.tdk.com...2ba503/ferrites-and-accessories-db-130501.pdf
EFD25 = 58mm^2 Page 546/547

Minimum turns to avoid saturation.

Nmin = LIpk/Bsat.Ae

Use short circuit current for Ipk... 15A

11 turns. The required Al value is 13E-6/11^2 or 107nH. You will have to use the formulae elsewhere to find an unequal gapped pair to achieve this. "Calculation factors (for formulas, see “E cores: general information”, page 402/403)"

Then you have to fit 11 turns on the bobbin in a way that makes sense and takes AC copper losses, skin/layer, into account. Same as you did for your transformer but, as was the case with core losses, the copper losses will be markedly reduced. 11 turns 6 by 0.5mm rope as a single layer?