 Consider a circuit consisting of a coil characterized by the inductance L and a resistor R connected in series. Then we take a voltage source, which provides a voltage U0, as soon as we close the circuit with a switch. Then a time-dependent current I flows through the coil and the resistor. The current does not have its maximum value immediately, but increases slowly due to Using Kirchhoff's laws, we can set up the following differential equation for the current I. Remember that the point above the I means the first time derivative. This is an inhomogeneous linear differential equation of first order. The searched function Y corresponds here to the current I. The perturbation function S corresponds to U0 over L and is time-independent in this case. We denote the homogeneous solution by IH. First we need to determine the homogeneous solution IH. We can quickly calculate this using the solution formula for the homogeneous version of the differential equation that you learned before. The coefficient K in front of the searched function I corresponds to R over L and is also time-independent in this case. So far so good. We may omit the constant CH in the solution formula here because we consider it later anyway in the constant A, which we find in the other solution formula. The coefficient R over L is constant and integrating a constant only introduces a variable T. Thus the homogeneous solution is IH is equal to E to the power of minus R over L times T. Let's insert it into the inhomogeneous solution formula. Note that one over the exponential function containing a minus in the exponent is simply equivalent to the exponential function without the minus sign. Now we have to calculate the integral. U0 over L is a constant and can be placed in front of the integral. And when integrating the exponential function, the exponential function is preserved. Simply L over R is added as a factor in front of the exponential function. Finally, we hide the constant of integration in the constant A. And this is our general solution. We can simplify it a bit more by multiplying out the parentheses. Two exponential functions simplify to one. To get a solution specific to the problem, we need to determine the unknown constant A. For that, we need an initial condition. If we say that the time T equals 0 is the time when the current I was 0 because we have not yet closed the switch, then our initial condition is y of 0 is equal to 0. Insert it into the general solution. e to the power of 0 is equal to 1. Solve for A and you get A is equal to minus U0 over R. Thus we have successfully determined the specific general solution.