Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.
Experiments show that at each instant a radioactive substance decomposes—and is thus decaying in time—proportional to the amount of substance present.
Step 1. Setting up a mathematical model of the physical process. Denote by y(t) the amount of substance still present at any time t.
By the physical law , the time rate of change y'(t) = dy/dt is proportional to y (t) this gives the first-order ODE.
(6) dy/dt = -kt
where the constant k is positive , so that , because of the minus , we do get decay.
The value of k is known from experiments for various radioactive substances (e.g., k=1.4/10^-11sec^-1 approximately , for radium )
Now the given initial amount is 0.5 g, and we can call the corresponding instant Then we have the
initial condition This is the instant at which our observation of the process begins. It motivates
the term initial condition (which, however, is also used when the independent variable is not time or when
we choose a t other than ). Hence the mathematical model of the physical process is the initial value problem
(7) dy/dt = -kt , y(0) = 0.5.
Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decay
and has the general solution (with arbitrary constant c but definite given k)
(8) y(t) = ce^-kt
We now determine c by using the initial condition. Since y(0) = c from (8), this gives y(0) = c = 0.5. Hence
the particular solution governing our process is (cf. Fig. 5)
(9) y(t) = 0.5 e^-kt
Always check your result—it may involve human or computer errors! Verify by differentiation (chain rule!)
that your solution (9) satisfies (7) as well as y(0) = 0.5:
dy/dt = -0.5ke^-kt = -k/0.5e^-kt = -ky , y(0) = 0.5e^0 = 0.5.
Step 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts from
the correct initial amount and decreases with time because k is positive. The limit of y as t → ∞ is zero.