Solve for B, t1, t2
A1 =B.e^pt1
A2=B.e^pt2 .
t2 - t1 = c ; known constant.
p also another known constant.
Any method is acceptable.
If A1=A2/(e^p.c) then the system has infinity number of solution, as
B=u, t1=ln(A1/u)/p, t2=t1+c.
If A1A2/(e^p.c) then the system has not a soluton.
If A1=A2/(e^p.c) then the system has infinity number of solution, as
B=u, t1=ln(A1/u)/p, t2=t1+c.
If A1A2/(e^p.c) then the system has not a soluton.
You can't find an explicit solution for this system, it is underdetermined. You have 5 unknowns - A1, A2, B, t1 and t2 and three equations. You probably need two more equations or try to substitute some pair of unknowns with parameter.
Greetings
Piotr
he said solve for B, t1, t2, which means i assume that you know A1 and A2, so you can do the following -
first note thatich has
A1 / A2 = t1 / t2,
which can almost always be solved using the other equation t2 - t1 = c,
so now we have t1 and t2,
so now A1 / t1 = A2 / t2 = B * e ^ p,
and since we know p,
we can solve for B
so the only issue is whether or not there is a solution to the combined equations (there are minor issues about 0, but if either A1 or A2 = 0, then B = 0 and t1 and t2 are arbitrary)
so t1 = t2 * (A1 / A2),
c = t2 - t2 * (A1 / A2),
which has a solution if A1 != A2 (if they are equal, then t1 = t2 and B is easy to find)
t2 = c / (1 - A1 / A2),
t1 = c / (A2 / A1 - 1)
or did i miss something?
perhaps it would be useful to know more about exactly what is assumed to be known at the start of the equation analysis
ok, so i did miss something - it appears that the t1 and t2 are in the exponent of their respective equations
in that case,
A2/A1 = e ^ (p * (t2 - t1)) = e ^ (p * c)
so if that condition does not hold,
then there are no solutions
if the condition does hold,
then
A1 = B * e ^ (p * t1)
A1 * e ^ (p * c) = A2 = B * e ^ (p * c + p * t1),
which now has indefinitely many solutions -
choose t1, compute B, compute t2
Afet Fatullayev has exactly the right answer
Given
A1 =B.e^pt1 (1)
A2=B.e^pt2 . (2)
t2 - t1 = c (3)
A1, A2, p, c are known. B, t1, t2 are unknown.
take natural log of (1) and (2)
ln(A1) = ln(B) +pt1 (4)
ln(A2) = ln(B) +pt2 (5)
(4) and (5) form the equation for a straight line with ln(B) as intercept and p as slope. Now the question boils down to finding the intercept and two values of the independent variable given the slope and the corresponding dependent values. Hence the straight line can be generalized as
ln(A)= ln(B)+pt
Hence the intercept ln(B) = ln(A) at t=t0 ( Since initial conditions are not given, it needs to be assumed and apply either to (4) or (5))
Now this further boiled down to a problem of linear equation given the intercept and the slope. Find either t1 or t2 since one can be deciphered from the other using (3)
t1=(ln(A1)-ln(B))/p
t2= t1+c
B=exp(ln(A)|t=t0)
given the following equations:
A1 = B*exp(p*t1) (1)
A2 = B*exp(p*t2) (2)
t2-t1 = c (3)
you may find the following solutions for t1 and t2:
t1 = (logA1-logB)/p (4)
t2 = (logA2-logB)/p (5)
from (3) it follows that:
c = (t2-t1) = (logA2-logA1)/p (6)
therefore:
logB=0
and it follows that:
B = 1 (7)
t1 = (logA1)/p (8)
t2 = (logA2)/p (9)
greetings
Dr Antonio Spano
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