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Calculating the steady state value is easy enough, it's just the duty cycle times the peak voltage:
\$ \dfrac{20ms}{60ms} \cdot 5V = 1.666V \$\$ \dfrac{20ms}{60ms} \cdot 5V = 1.\overline{666}V \$
\$ \dfrac{20ms}{60ms} \cdot 5V = 1.666V \$
\$ \dfrac{20ms}{60ms} \cdot 5V = 1.\overline{666}V \$
