"why is the voltage increase with the duty cycle?"
Simple: if you increase the DutyCycle, you charge the inductor for longer, hence it contains more energy at the end of the charge. On the discharge cycle this energy is transferred via diode D into the load and the capacitor.
Yes ILmax will be higher because it will be charged more. Realize that a capacitor is charged with current and the voltage shows how much it is charged. With an inductor it's the other way round, it is charged by applying a voltage (this happens when S closes) and the current shows how much it is charged.
You focus a bit much on Vl, the voltage across the coil. But that is not so important, the inductor current is what matters. The inductor behaves as a current source when it still contains charge and S is open.
Note that for the same dutycycle if you increase the value of the load R, the output voltage V0 will increase ! A given Dutycycle does not result in a constant voltage at the output. So that's why real boost converters need a feedback circuit to control the dutycycle.
If you would remove load R, the voltage would increase to infinity ! (in theory that is)
The continuous mode boost and buck-boost converters exhibit control-to-output transfer functions \$G_{vd}(s)=\hat{v}(s)/\hat{d}(s)\$ containing two poles and one RHS (right half-plane) zero, called zero of nonminimum phase.
Your original transfer funtion is:
$$G_{vd}(s)=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C\times\left ( 1-s\left ( \underline{L}/R \right ) \right )}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$
Starting from a simpler function (without the RHP zero), named \$ G_{vd}'\$:
$$G_{vd}'=\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$
Or, placed as a standard second order t.f:
$$G_{vd}'=K_{DC}\times \frac{1/\underline{L}C}{s^2+s\left ( 1/RC \right )+1/\underline{L}C}$$
where the DC gain is \$K_{DC} =\frac{1}{V_{RAMP}}\times \frac{V_{IN}}{\left ( 1-D \right )^2}\$.
The equation can be rewritten as:
$$G_{vd}'=K_{DC}\times \frac{1}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$
With \$\omega_0=1/\sqrt{\underline{L} C}\$ and \$\omega_0Q=R/\underline{L}\$
Similarly, the original transfer function can be expressed as (RHP zero included):
$$G_{vd}=K_{DC}\times \frac{\left (1-\frac{s}{\omega_{RHP}} \right )}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }$$
The response will be:
$$\hat{v}(s)= \frac{K_{DC}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1}\times\hat{d}(s) - \frac{K_{DC}/\omega_{RHP}}{\left ( \frac{s}{\omega_0} \right )^2 + \frac{s}{\omega_0Q}+1 }\times\hat{d}(s)\times s$$
The response of the original system is the sum of two components: The first is equivalent to the modified system response (without the zero) and the second is the derivative (scaled) of that one. For the case of a stable system with a step input in \$t = 0\$, this last component will have substantial influence at the beginning and then will vanishes when \$t\rightarrow\infty \$. Note the negative sign leads to a momentary opposite effect in output (nonminimum phase).
UPDATE:
The presence of zero RHP in the model is explained as follows: For the output voltage to increase, the duty cycle must be increased in such a way that the inductor will be disconnected from the load for a long time, causing the output voltage to drop (i.e. in the opposite direction to the one desired). The controller must be designed to meet the project requirements and avoid oscillations while maintaining duty cycle below an undesirable 100% - being limited by the PWM integrated circuit itself.
Best Answer
It stays constant because the incoming supply defines that voltage but, if you left the inductor connected across the supply, after a few tens of microseconds, the current might be tens or hundreds of amps and that will cause the incoming supply to collapse. Then you're in trouble.
But that isn't how a boost circuit works. The inductor is placed in parallel with the supply and current ramps up to "several" amps. That peak current is, in effect, representing the stored energy in the inductor. Microseconds later, the inductor is disconnected from the incoming supply and, that stored energy is released into the output capacitor and load. Then the process starts again.
The point is that you don't hold the inductor connected to the incoming supply for more than a few microseconds (or a few tens of microseconds in really powerful boost converters). Here's an example of a boost converter's inductor current: -
Web boost-regulator calculator.
On the example above, current ramps up from 0 amps to 15.656 amps in 6.132 μs. It then transfers the energy (during which it ramps down) in 2.044 μs and, a little while later (1.824 μs), the cycle begins again. This is just one example and, we'd expect the incoming supply to be able to deliver 15.656 amps without drooping too much. This boost converter is operating in discontinuous conduction mode (\$\color{red}{\boxed{\text{DCM}}}\$) because the inductor current falls to zero before the cycle restarts.
A larger value inductor (10 μH from 4.7 μH) means the current peak is less: -
Peak current is 10.9 amps and now the switching circuit is operating in what is known as continuous conduction mode (\$\color{blue}{\boxed{\text{CCM}}}\$) where the inductor holds on to a little bit of energy as the cycle starts again