First Order Differential Equations


Example: Find the solution to the following differential equation.

Solution:

First we need to get the differential equation in the correct form.

From this we can see that p(t) = 0.196 and so μ(t) is then.

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.

We can subtract k from both sides to get.

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 e0.196t or to multiply both sides by e−0.196t . Either will work, but I usually prefer the multiplication route. Doing this gives the general solution to the differential equation.



Charging A Capacitor With Constant Current

A capacitor 1F is being charged by a constant current I. Find the voltage vcacross this capacitor as a function of time given that the voltage at some reference time t = 0 is Vo ic(t) = constant

C = 1F

By separation of the variables,

and by integrating both sides of we get



RL Circuits

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, VL 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
xO                      0

ln ( x / xO ) = - ( R / L ) t

where xO is the value of x at time t = 0

Taking the antilogarithm:

x = xO e - ( R t ) / L

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

  V          V
----- - I = --- e - ( R t ) / L
  R          R

      V
I = --- ( 1 - e - ( R t ) / L )
      R







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