Example 2

A radioactive element has a half-life of N years, that is, half of the substance will decay every N years. Given ten pounds of the element at time t = 0, how much will remain at time r? In radioactive decay the amount y of the element is decreasing at a rate proportional to y, so the differential equation has the form

dy/dt = ky.

The general solution is

y = Ce^kt.

Since y is decreasing, k will be negative. We must find the constants C and k. To find C we use the initial condition

y = 10 at f = 0, C = 10.

To find k we use the given half-life. It tells us that y = ½ · 10 = 5 at f = N.

Therefore

The solution is

y = 10e^{-(t ln 2)/N}.