Ideally, the output voltage of a buck converter running in continuous mode is a multiple of the input voltage, with that multiple dependent only on the duty cycle.
However, back here in the real world it's hard to find those ideal diodes with no forward voltage drop, inductors with no series resistance, and capacitors with no losses. The various non-idealities cause some voltage drop with higher current.
The usual solution is to close the loop to control the duty cycle to whatever it takes to achieve the desired output voltage. With good parts, that duty cycle will remain largely constant for a particular combination of input and output voltage. However, it will go up some with higher output current. That's because the control loop has to push the circuit a little harder to compensate for the inevitable losses.
For a boost converter, it's more complicated. Unlike with a buck converter, the duty cycle is a tradeoff between enough time to store energy in the inductor, and enough time to deliver the stored energy to the load. 100% duty cycle, for example, continuously charges up the inductor but never delivers anything to the output.
However, there is still a fixed voltage ratio between input and output that is only a function of the duty cycle with ideal components. For a buck converter, if D is the fraction of the time the inductor is connected to the input voltage and the inductor is connected to ground the remaining time (1-D), then the ratio of output voltage to input voltage is simply D.
Vout / Vin = D
Now think of a boost converter as the buck converter run in reverse. That means Vout and Vin are swapped. It also means with consider the "on" time of the inductor when it is connected to ground, not Vout. Therefore D of a boost converter is 1-D of the same thing viewed as a buck converter.
Applying all this flipping from buck to boost to the equation above, yields the equation for a boost converter:
Vin / Vout = 1 - D
Rearranging this to tell us what the output to input voltage ratio is yields:
Vout / Vin = 1 / (1 - D)
This is easier to see by analyzing a simplified switching converter:
First let's consider this a buck converter. You stipulated continuous mode, so the switch always either connects the left side of the inductor to VA or ground. The result is a simple low pass filter.
However, this same circuit works in reverse as a boost converter. With the switch always connected to one of the two choices and no diode, this is really a DC transformer. It works the same way for either buck (input is VA, output is VB) or boost (input is VB, output is VA). If we consider the duty cycle to be the fraction of time that the switch is connected to VA, then VB is simply the duty cycle times VA. That's the buck converter view.
The relationship works identically in reverse. VA is VB divided by the duty cycle. The only difference for typical boost converter analysis is that we usually consider the duty cycle the fraction of the time the switch is connected to ground instead of VA. In other words, we use 1-D relative to what we call "duty cycle" for a buck converter.
Now before you complain that this is unfair because the diode is missing and that current can run backwards thru the inductor, remember that you stipulated the converter was running in continuous mode. The duty cycle, input and output voltage ratios, and output current demand are such that the current is always flowing in the inductor. If you knew this was always true, you could remove the diode.
The circuit as shown, without a diode, does actually work both ways, and inductor current can flow either direction. This is basically a "DC transformer", with the voltage ratio strictly a function of the duty cycle, no matter which way you define it.