Parallel RL-Circuit with non-sinusoidal current source

Question

Consider a parallel RL-circuit, connected to a current source $I(t)$. The signal is for the moment arbitrary, so not sinusoidal.

QUESTION 1: Find the differential equation of the voltage $U(t)$.

In the case of a RL (or RLC) circuit in series one would write: $$ U(t) = \frac{1}{C}\int{I(t)dt}+L\frac{dI(t)}{dt}+RI(t) $$ In parallel - in our case - one can write: $$ I(t) = \frac{1}{R}U(t)+\frac{1}{L}\int{U(t)dt} $$ but I don't know how to express $U(t)$.

QUESTION 2: Calculate $U(t)$ for

$$ I(t) = \begin{cases} 0 & t<0\\ I_0 & t\geq 0\\ \end{cases} $$

How would I approach these two questions?

Answer 1

After Alfreds hint for Q1: $$ U(t) = L\frac{dI(t)}{dt}-\frac{L}{R}\frac{dU(t)}{dt} $$ For Q2: $$ U(t) = L\frac{d}{dt}\left(I_0 \Theta(t)\right)-\frac{L}{R}\frac{dU(t)}{dt} = LI_0 \delta(t)-\frac{L}{R}\frac{dU(t)}{dt} $$ Solving the differential equation: $$ U(t) = C_1 e^{-\frac{Rt}{L}}+I_0 R \Theta(t) e^{-\frac{Rt}{L}} $$ with an unknown constant $C_1$.

The current through an inductor $L$ must always be continuous, otherwise it would defy the conservation of energy. Because $I(t)$ is discontinuous at $t=0$, the "jump" of current flows solely through the resistor $R$. So we see: $$ U(t=0) = RI_0 $$ It then follows that $C_1=0$ and $$ U(t) = I_0 R \Theta(t) e^{-\frac{Rt}{L}} $$

Why it makes sense: The voltage of an inductor $U_L=L\frac{dI}{dt}$ is dependend on the change of current. With no change for a long time ($t\rightarrow \infty$) $U_L=0$, the inductor is a short circuit.