| The integral in .0 is actually a bit more general than it needs to be.
The original differential equation is:
(1) L2*I0/(k*T)*exp( V/[k*T] ) * (dV/dt)
+ R2*I0*[ 1 - exp( V/[k*T] ) ] - V = 0
V is the dependent variable representing the voltage across a diode
as a function of the independent variable t (time).
L2, I0, k, T, and R2 are constants.
exp(x) denotes the exponential function 'e to the x'.
The need for the integral arises from trying to find an integrating
factor for eqn (1).
If eqn (1) is matched against M(x,y) + N(x,y)*(dy/dx), then
y ~ V
x ~ t
(2) M(x,y) ~ R2*I0*[ 1 - exp( V/[k*T] ) ] - V = M(t,V)
(3) N(x,y) ~ L2*I0/(k*T) * exp( V/[k*T] ) = N(t,V)
where '~' means 'corresponds to'
There is a theorem (see 'Elementary Differential Equations and Boundary
Value Problems', 3rd ed., Boyce & DiPrima, sec. 2.6, p. 44, prob. 10)
that:
If (Nx - My)/M = Q with Q a function of y only, then
M + N*(dy/dx) = 0 has an integrating factor of the form
u(y) = exp( indefinite_integral( y, Q ) )
where Nx and My denote the partial derivatives of M and N
w.r.t. x and y respectively.
By inspection of (2) and (3) it can be seen that the correspondent of
Q must be a function of V only (since M(t,V) and N(t,V) in this case are
functions of V only).
Computing Q:
Nx ~ 0
My ~ - R2*I0/(k*T)*exp( V/[k*T] ) - 1
M ~ R2*I0*[ 1 - exp( V/[k*T] ) ] - V
therefore
(4) -My/M ~ Q(t,V) = ( a + b*exp( V/a ) ) / ( a*b*[ 1 - exp( V/a ) ] - a*V )
with a = k*T and b = R2*I0 to improve readability
Seeking a solution for (4) was the basis for the note in .0.
The form of .0 is actually more general than that of (4).
By happy coincidence, Q(t,V) is an exact differential of:
(5) integral_Q = - ln( a*b*[ 1 - exp( V/a ) ] - a*V )
This is as far as I've gotten.
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