Linear First-Order Ordinary Differential-Equation

This problem does not fit into any place; but, I am asked this question often enough that I may as well post it *someplace*.
We will discuss the first-order ordinary differential-equation in terms of a series RC or LR circuit. The inductor, of inductance L, the resistor, of resistance R, and the condenser, of capacity C, are in series. We will write the equation in terms of the charge q or current i, as a function of time t.

Theorem:  The linear first-order ordinary differential-equation

• y' + f(x) y = g(x)

has the general solution

• y = (y(0) + integral, from 0 to x, of (h(x) g(x) dx)) / h(x),

where the integrating factor h(x) is defined as

• h(x) = exp(integral, from 0 to x, of f(x)).

Proof:  Now, multiply the differential equation by the integrating factor h(x), to obtain

• h(x) y' + h(x) f(x) y = h(x) g(x).

Integrate it, to obtain

• h(x) y(x) = h(0) y(0) + integral, from 0 to x, of (h(x) g(x) dx).

Observe that h(0) = 1.  Divide by h(x), to obtain

• y(x) = (y(0) + integral, from 0 to x, of (h(x) g(x) dx)) / h(x).

QED.  There you have it.

Numerical examples

the function f is a constant

series RC circuit

The resistor, of resistance R and the condenser, of capacity C, are in series.  The differential equation for the charge q as a function of time t is

• R q' + q / C = e(t),

where e is the driving function.

Divide by R, to obtain

• q' + (1 / (R C)) q = e(t) / R.

By comparison, the functions f and g are

• f(t) = 1 / (R C)
• g(t) = e(t) R.

Substitute into the general solution, to obtain

• h(t) = exp(integral, from 0 to t, of ((1 / (R C)) dt)) =
  = exp((1 / (R C)) t)
• q(t) = (q(0) + integral, from 0 to t, of ((exp((1 / (R C)) t) e(t) / R) dt).

The special case of constant e(t) = e yields

• q(t) = (q(0) + (C e) (exp((1 / (R C)) t) - 1) exp(- (1 / (R C)) t) =
  = (q(0) - (C e)) exp(- (1 / (R C)) t) + C e.

Its derivative is the current i

• i(t) = q'(t) = - ((q(0) - (C e)) / (R C)) exp(- (1 / (R C)) t) =
  = - ((q(0) / (R C)) - (e / R)) exp(- (1 / (R C)) t).

At the initial time t = 0, they are

• q(0) = q(0)
• i(0) = q'(0) = - (q(0) / (R C)) - (e / R).

At the time constant t = R C, they are

• q(R C) = (q(0) - (C e)) exp(- 1) + C e
• i(R C) = q'(R C) = - (q(0) - (C e)) exp(- 1).

Their limits as t increases without bound are

• q(oo) = C e
• i(oo) = q'(oo) = 0.

The potential drops across each of the two components are:

• across C
  q / C = ((q(0) / C) - e) exp(- (1 / (R C)) t) + e. 
• and across R
  q' R = - ((q(0) / C) - e) exp(- (1 / (R C)) t).

Their sum, which should equal the driving function, is

• e.

series LR circuit

The inductor, of inductance L, and the resistor, of resistance R, are in series.  The differential equation for the current i as a function of time t is

• L i' + R i = e(t),

where e is the driving function.

Divide by L, to obtain

• i' + (R / L) i = e(t) / R.

By comparison, the functions f and g are

• f(t) = R / L
• g(t) = e(t) R.

Substitute into the general solution, to obtain

• h(t) = exp(integral, from 0 to t, of ((R / L) dt)) =
  = exp((R / L) t)
• i(t) = (i(0) + integral, from 0 to t, of ((exp((R / L) t) e(t) / R) dt).

The special case of constant e(t) = e yields

• i(t) = (i(0) + (e / R)) (exp((R / L) t) - 1) exp(- (R / L) t) =
  = (i(0) - (e / R)) exp(- (R / L) t) + e / R.

Its derivative is the current i

• i'(t) = - ((i(0) - (e / R)) (R / L)) exp(- (R / L) t) =
  = - ((i(0) R / L) - (e / L)) exp(- (R / L) t).

At the initial time t = 0, they are

• i(0) = i(0)
• i'(0) = - (i(0) R / L) - (e / L).

At the time constant t = L / R, they are

• i(L / R) = (i(0) - (e / R)) exp(- 1) + e / R
• i'(L / R) = - (i(0) R / L - (e / L)) exp(- 1).

Their limits as t increases without bound are

• i(oo) = e / R
• i'(oo) = 0.

The potential drops across each of the two components are:

• across R
  i R = (i(0) R - e) exp(- (R / L) t) + e.
• and across L
  i' L = - (I(0) R - e) exp(- (R / L) t).

Their sum, which should equal the driving function, is

• e.

Each of these examples has both the f(x) and g(x) constant. If someone will suggest some interesting examples of variable f(x) or g(x), I will work them out and post them here.

I did attempt to be careful. I believe that the foregoing is correct. However, I am notoriously error-prone. Hence, beware! :-) If you should discover any errors or if you have comments or suggestions, please contact me. Grammercy.

Copyright (c) 2001 by R.I. 'Scibor-Marchocki. Webmaster@rism.com Last modified Friday 20-th July 2001.