Final answer:
The differential equation for a circuit with a unit step voltage as a function of time relates the time rate of change of energy in a capacitor to the product of the voltage across it and the rate of change of the voltage, which is represented by dU/dt = C*v(t)*dv(t)/dt. It's a first-order differential equation that can be integrated to find the charge on the capacitor as a function of time.
Step-by-step explanation:
Writing the Differential Equation for a Circuit with a Unit Step Voltage
To write a differential equation for a circuit when v(t) equals u(t), a unit step, we first consider the energy inside a capacitor. The time rate of change of energy in a capacitor is related to the voltage across the capacitor, V(t), and the rate of change of voltage, dv(t)/dt. For a capacitor, the equation for the energy U stored at any time t is U = (1/2)C[v(t)]^2, where C is the capacitance. Thus, the rate of change of energy with respect to time can be expressed as dU/dt = C*v(t)*dv(t)/dt.
This differential equation represents how the charge on the capacitor evolves over time and is a first-order differential equation for the current I(t). It has a notable similarity to the equation for an RC circuit, where a resistor and capacitor are in series. To obtain the equation for the charge on the capacitor as a function of time, we would integrate this equation with the given boundary conditions.
By integrating the differential equation, starting from when the unit step is applied (at t = 0), the charge on the capacitor can be found. From these considerations, you can write the differential equation for the circuit response to a unit step input of voltage.