Section 6-2-2

Chapter 6
Antidifferentiation

6.2 Separation of Variables

6.2.2 Symbolic Solutions 1: Natural Growth

Before we look for a symbolic solution for the constrained growth equation, we will explore another way to solve the natural growth equation,

\frac{d P}{d t} = k P,

one that also will apply to constrained growth. When we first solved this equation in Chapter 2, we guessed the answer and checked by direct calculation that the answer is correct. That won't work with constrained growth, though, because we have no information about a likely form for the answer.

The derivative `(dP)/(dt)` may be considered as the quotient of the differentials `dP` and `dt`. Recall that the differential of the dependent variable `P` is just

d P = \frac{d P}{d t} d t .

Now multiply both sides of the differential equation `(dP)/(dt)=kP` by the differential `dt`:

d P = k P d t .

Divide both sides of this equation by `P` to find

\frac{d P}{P} = k d t .

The right side of this equation is the differential of

k t + C,

where `C` can be any constant. We'll try to write the left side, `(dP)/P`, as a differential as well.

For the moment, think of `P` as an independent variable. Then `(dP)/P` is the differential of any function whose derivative is `1/P`. One such function is `lntext[(]Ptext[)].` [We know that `P` is positive, so it actually has a (natural) logarithm.] But is `lntext[(]Ptext[)]` an antidifferential for `(dP)/P` when we consider `t` as the independent variable? Let's check. We know that

d \ln (P) = \frac{d}{d t} \ln (P) d t .

By the Chain Rule,

\frac{d}{d t} \ln (P) = \frac{1}{P} \frac{d P}{d t} .

$d \ln (P)$	$= \frac{1}{P} \frac{d P}{d t} dt$
	$= \frac{1}{P} d P .$

Here is the important point: For antidifferential calculations, the Chain Rule shows that it doesn't make any difference whether we think of `P` as an independent variable or a dependent variable.

So for the function `P` that we seek, the differential of `lntext[(]Ptext[)]` equals `k dt`. Thus `lntext[(]Ptext[)]` must be one of the functions

k t + C .

Now we use the initial condition to determine which constant `C` we need. Since `Ptext[(]0text[)]=P_0`, we know that

\ln (P_{0}) = 0 + C,

C = \ln (P_{0}) .

Thus

\ln (P) = k t + \ln (P_{0}) .

Finally, we get rid of the logarithms by taking exponentials on both sides:

P = e^{\ln (P)} = e^{k t + \ln (P_{0})} = e^{k t} e^{\ln (P_{0})} = P_{0} e^{k t} .

Not surprisingly, we get the same answer we found in Chapter 2.

Contents for Chapter 6

Chapter 6 Antidifferentiation

6.2 Separation of Variables

Chapter 6
Antidifferentiation