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In an RL circuit, we often deal with a first-order linear differential equation. This type of equation involves the first derivative of the current. For example, in the exercise, the RL circuit differential equation is: equation: \[ L \frac{d i}{d t} + R i = u \] L represents the inductance, R the resistance, and u the voltage across the circuit. These equations are named 'first-order' because they contain the first derivative of the current (\( \frac{d i}{d t} \)). Moreover, they are 'linear' because the current and its derivative appear linearly (i.e., they aren't raised to any power other than one).Understanding this type of differential equation is fundamental, as they describe how current changes over time in an RL circuit.

To solve a first-order linear differential equation like the one in our RL circuit, we must split it into two parts: the homogeneous solution and the particular solution.The homogeneous solution addresses the differential equation when the driving voltage, \( u(t) \), is zero. For our problem, it starts with:\[ L \frac{d i}{d t} + R i = 0 \]The solution to this homogeneous differntial equation is\[ i_h(t) = C e^{-\frac{R}{L} t} \]Here, 'C' is an integration constant determined later using initial conditions.Next, we find the particular solution, which addresses the specific form of the driving function. For constant voltage \(u(t) = u_0\), our particular solution becomes a constant as well. Solving for when \( \frac{d i_p}{d t} = 0 \):\[ R i_p = u_0 \implies i_p = \frac{u_0}{R} \]Thus, the current in our RL circuit is a combination of the homogeneous solution and the particular solution: \[ i(t) = i_h(t) + i_p = C e^{-\frac{R}{L} t} + \frac{u_0}{R} \]

Initial conditions help us determine the unknown constants in our solutions. In our problem, the initial condition given is \(i(0) = 0\). Using this, we plug it into our general solution for part (a):\[ 0 = C e^{-\frac{R}{L} \cdot 0} + \frac{u_0}{R} \implies 0 = C + \frac{u_0}{R} \implies C = -\frac{u_0}{R} \]Thus, the complete current solution becomes:\[ i(t) = \frac{u_0}{R} \left( 1 - e^{-\frac{R}{L} t} \right) \]Initial conditions in differential equations always serve to pinpoint the unique solution that corresponds to the specific scenario under consideration.

Sometimes, the voltage in an RL circuit is not constant but changes with time. For instance, if the voltage increases linearly with time \(u(t) = at\), our approach in solving the differential equation remains similar but a bit more complex. For the given scenario:\[ L \frac{d i}{d t} + R i = at \]We again find both homogeneous and particular solutions.The homogeneous part remains the same:\[ i_h(t) = C e^{-\frac{R}{L} t} \]For the particular part, assume a form \( i_p(t) = At + B\) and substitute to find A and B:\[ A R = a \quad \text{and} \quad R B + L A = 0 \]Solve the above to find A and B:\[ A = \frac{a}{R} \quad \text{and} \quad B = -\frac{a L}{R^2}\]The complete solution then becomes:\[ i(t) = C e^{-\frac{R}{L} t} + \frac{a}{R} t - \frac{a L}{R^2} \]Apply the initial condition, and we get:\[ C = \frac{a L}{R^2} \]Thus, the solution for linearly increasing voltage is:\[ i(t) = \frac{a L}{R^2} \, e^{-\frac{R}{L} t} + \frac{a}{R} t - \frac{a L}{R^2} \]