If a quantity grows at a rate which is proportional to its size (such a in populations of bacteria, or interest rate, etc.) it will satisfy a relation of the form

dy/dt = k y(t)

where k is constant. If k > 0, y(t) increases. If k < 0, then y(t) decreases. There is only one kind of function which satisfies this relation, the exponential function, y(t) = C e^kt.

To see why this is so, rewrite the relation as

1/y(t) dy/dt = k, or
d/dt (ln (y(t))) = k

which implies that d/dt (ln(y(t))-kt) = 0, which means that ln(y(t))-kt = C (constant)! Solving for y(t) yields ln(y(t)) = kt+C, and therefore y(t) = e^kt+C.

Theorem: The only solutions of the differential equation dy/dt = ky are the exponential functions

y(t) = y(0) e ^kt

Given two values of y(t) (at two different times) means we can solve for k, the growth constant...

y1 = y(t₁)=y(0) e^kt₁ and y2 = y(t₂)=y(0) e^kt₂. Dividing, we have y1/y2 = e^k(t₁-t₂)

Taking the natural log of both sides, ln(y1)-ln(y2) = k(t₁-t₂), therefore k=( ln(y1)-ln(y2) )/(t1-t2)!

Knowing y(t) at two later times t1,t2 > 0 also means we can find the initial amount y(0).

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As a practical example of exponential growth/decay, consider Newton's Law of Cooling , which states that the rate of temperature cooling is proportional to the temperature difference

dy(t)/dt = k(y(t)-A),

where A=ambient temperature, and k is negative.

If we let u(t)=y(t)-A, this reduces to the previous relation

du(t)/dt = k u(t) This means y(t) - A = u(t) = u(0) e ^kt , and therefore y(t) = A + u(0) e ^kt .

Another nice example, that of compound interest, is given in the text.

Last modified Wed Oct 23 23:19:00 CDT 1996