Absolute values in logarithms in a solution of differential equationHow have the moduli signs disappeared...

Araceli Clay

Araceli Clay

Answered

2022-07-03

Absolute values in logarithms in a solution of differential equation
How have the moduli signs disappeared in the following step:
1 k ( ln | g + k v | ln | g + k u | ) = t
Therefore
ln ( g + k v g + k u ) = k t
g, k and u are positive constants. t is time, v is velocity.
Context: the above calculations are from solving the equation d v / d t = g k v given that v = u when t = 0, and that u, g and k are positive constants.

Answer & Explanation

gutinyalk

gutinyalk

Expert

2022-07-04Added 11 answers

g + k v need not be positive, but it will have the same sign as g + k u, because the solutions cannot cross the equilibrium at v = g / k. Hence, the quotient inside the logarithm is positive.
Ideally, you would not arrive at 1 k ( ln | g + k v | ln | g + k u | ) = t at all; there is a cleaner way of solving this ODE. Namely, one goes from ln | g + k v | = k t + C (with indefinite constant C) to g + k v = A e k t , where A takes the place of ± e C
rzfansubs87

rzfansubs87

Expert

2022-07-05Added 5 answers

If k, g, u
are all positive, you need to have v large enough (i.e., not too negative) to make g + k v > 0. Maybe that results from some part that you haven't told us...

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?