Why does the current rise and fall linearly in a DC-DC converter?
Most real switching supplies will have a capacitor in parallel with the load:







simulate this circuit &ndash; Schematic created using CircuitLab

It is this capacitor that makes the voltage supplied to the load approximately constant even with the switching. In the periods where the switch is closed and the inductor is not supplying current to the load, the load draws current by discharging C1.

Also keep in mind, there is usually something regulating the duty cycle of the switch to keep the output voltage, and thus the voltage across C1, at some target.

The graph in your question it graphing current and voltage for the inductor, not the load. When the switch is closed, the voltage across the inductor is V1. When the switch is open, the voltage across the inductor is the output voltage, plus a little more to forward-bias D1, minus V1.

Either way, there's a constant voltage across the inductor in either state. A constant voltage across an inductor results in a linearly changing current, according to the definition of inductance:

$$ v(t) = L{\mathrm di \over \mathrm dt} $$