Electronic – use digital pots for the feedback resistors to a buck-boost regulator

You may not need a power supply at all, just a diode between the solar panel and the battery. Lead-acid batteries can be float-charged for long periods of time without damage. If the maximum current the panel puts out at about 13.6 V doesn't exceed the float charge current of your battery, then there is no need for any converter between the panel and the battery.

If you're worried about efficiency, save the time, effort, and money that would go into the switching power supply and get a bigger solar panel instead. Sunlight is free, so efficiency is more relevant measured in effort or dollars per Watt than percentage of energy conversion. Get a solar panel just big enough to deliver the maximum float current at 13.8 V, then put only a Schottky diode between the panel and the battery.

Although the absolute value of a digital potentiometer's resistance may vary 30%, the matching between the internal resistors is really good. This means that for a voltage divider (to wit, a potentiometer), the voltage division accuracy is quite good, since the voltage division factor relies entirely on the ratio of the resistors used, not their absolute value.

If your output voltage is less than the digital potentiometer's maximum voltage rating, you can simply use a digital potentiometer to stand in for the feedback network without any fanfare. Most digital pots can take only 5.5V, but some are rated for substantially more.

If your output voltage is higher than the digital potentiometer's maximum rating, or if you want fine adjustment, you can combine the digital potentiometer with external resistors to form a compound voltage divider. Note that this will cause the digital potentiometer's absolute value variation to come into play. Techniques exist to minimize this error, as covered here.

If you want to avoid digital potentiometers altogether, you can make the feedback divider take input from a D/A as well, as shown here by having a D/A drive \$V_{CTRL}\$:

^{simulate this circuit – Schematic created using CircuitLab}

Since the feedback voltage \$V_{FB}\$ will be regulated by the LTC3780 to 0.8V, the output voltage will be regulated to:

$$V_O=\left(1+\frac{R_2}{R_1}\right)0.8-\frac{R_2}{R_3}(V_{CTRL}-0.8)$$

Setting \$V_{CTRL}\$ to 0.8V causes no change in the output voltage; increasing \$V_{CTRL}\$ causes the output to drop, and decreasing \$V_{CTRL}\$ causes it to rise.

It should be noted that no matter what you do, you should carefully evaluate the regulator and adjust components (\$L\$, \$C_{OUT}\$, \$I_{TH}\$ network) to ensure stability over all operating conditions. When in doubt, lean to the conservative side here.