d=dmax (1-e^(-kt))
This equation is the result of integrating an ordinary differential equation. Derive that ODE and the associated initial condition.
Use the equation to calculate dd(t)/dt, and express it in terms of d(t).
dd/dt = -k*dmax e^-kt = -k*(d - dmax)
Let d' = dmax - d
dd'/dt = -kd'
d' + (1/k) dd'/dt = 0
d' at t= 0 = dmax is the initial condition.
thank you...I was able to work through all the steps with your help!
To derive the ordinary differential equation (ODE) and the associated initial condition for the given equation d = dmax (1 - e^(-kt)), we can start by considering the rate of change of d with respect to time, which can be written as:
ddt = dmax * (-e^(-kt)) * k
Now, let's analyze this equation further. The derivative of d with respect to t represents the instantaneous rate of change of d at any given time, which can be interpreted as the rate at which the quantity d is changing. In this case, d represents some quantity that is increasing over time.
Now, the equation d = dmax (1 - e^(-kt)) suggests a scenario where d is gradually approaching a maximum value, dmax, under the influence of a constant, k. This indicates that the rate of change of d is proportional to the difference between dmax and d.
Based on this observation, we can write the ODE that describes the behavior of d over time as follows:
ddt = k * (dmax - d)
To understand the initial condition associated with this ODE, we need to know the value of d at a given initial time, t0. Let's denote that initial value of d as d0.
Therefore, we have the initial condition: d(t0) = d0.
So, the derived ODE and the associated initial condition are:
ODE: ddt = k * (dmax - d)
Initial Condition: d(t0) = d0
These equations represent the mathematical model that describes the behavior of the quantity d over time, and the initial condition specifies the value of d at the initial time.