Note that officially there should be a constant of integration in the exponent from the integration.

However, we can drop the constant.

Now multiply all the terms in the differential equation by the integrating factor and do some simplification.

Integrate both sides and don't forget the constants of integration that will arise from both integrals.

Both *c* and *k* are unknown constants and so the difference is also an unknown constant. We will
therefore write the difference as *c*. So, we now have

From this point on we will only put one constant of integration down when we integrate both sides knowing that if we had written down one for each integral, as we should, the two would just end up getting absorbed into each other.

The final step in the solution process is then to divide both sides by *e ^{0.196t}* or to multiply both
sides by

A capacitor *1F* is being charged by a constant current *I*. Find the voltage v_{c}across this capacitor as
a function of time given that the voltage at some reference time t = 0 is V_{o}
*i _{c}(t) = constant*

*C = 1F*

By separation of the variables,

and by integrating both sides of we get

Consider the circuit shown above, which contains a battery of small
internal resistance and inductor connected in-series with the battery. This is an **RL circuit** because the elements connected to the
battery are a resistor and an inductor. Suppose that the switch S is open for t < 0 and
then closed at t = 0. The current in the circuit begins to increase, and a back emf
that opposes the increasing current is induced in the inductor. Because the
current is increasing, di/dt is positive; thus, V_{L} is negative.
This negative value reflects the decrease in electric potential that occurs in going from a to b
across the inductor, as indicated by the positive and negative signs in the Figure above.

With this in mind, we can apply Kirchhoff’s loop rule to this circuit, traversing the
circuit in the clockwise direction:

**
di**

** V - IR - L --- = 0 **

**
dt**

where *IR* is the voltage drop across the resistor.
(We developed Kirchhoff’s rules for
circuits with steady currents, but they can also be applied to a circuit in which the
current is changing if we imagine them to represent the circuit at one instant of time.)
We must now look for a solution to this differential equation.

A mathematical solution of the Equation shown above represents the current in the circuit as
a function of time. To find this solution, we change variables for convenience, letting

**x = ( V / R ) - I**

so that **dx = - dI**

With these substitutions, we can write

** L dx**

**x + --- ----- = 0**

** R dt**

** dx R**

**----- = - ---- dt**

** x L**

Integrating both sides:

** x** t

** ∫ dx / x = - R / L ∫ dt**

**x _{O} 0**

where x

Taking the antilogarithm:

Because I = 0 at t = 0, we note from the definition of x that x_{O} = V/R. Hence, this last
expression is equivalent to

** V V**

**----- - I = --- e ^{- ( R t ) / L}**