If an organism reproduces without competition from other organisms in an environment which can support up to L organisms (L is called the carrying capacity of the environment), then we expect that the population of the organism will tend to slow as the population approaches this limit.

• If the population of the organism is P, representing a function P(t) of time, then the rate which the organism grows is dP / dt.
• We expect that this rate dP / dt will decrease as P -> L.
• We wish to set up a model by speculating to what dP / dt should be proportional.

The figure below depicts the expected way the population will approach L.

• Note that dP / dt, the slope of the graph, is decreasing as P -> L.

As an example of a proportionality for dP / dt, we try dP / dt = k P.

• The solution, which should be familiar, is indicated below.
• However, we do not expect that the population will increase at a faster rate when P increases; rather we expect that the population will increase more more slowly as P increases toward L.

We speculate that perhaps dP / dt is proportional to the difference L - P, which we see is also decreasing as P -> L.

• We note that this proportionality should not be valid for small populations P < < L, which we expect to increase more and more quickly due to greater and greater numbers of reproducing organisms.

• We do not expect that the rate of population increase will begin to level off until the population gets close to L.

If dP / dt is proportional to L - P, then dP / dt = k ( L - P ) for some k.

• We write this proportionality as a differential equation and proceed to solve by separating the variables, as indicated below.

The last few lines of the solution, and a graph of the resulting population function, are shown below.

The function P(t) = L - C e^(-kt), for a given value of k, will give us a family of solutions for different values of C.

We look first at the functions C e^(-kt).

• For positive values of C we have the family C e^(-kt) indicated below by C = 4, ..., C = 1, approaching the t axis from above.
• For negative value of C we have the function approaching the t axis from below.
• The half-life of the function is the same for every value of C.

The function P(t) = L - C e^(-kt) will therefore have the family indicated in the figure below, where the exponential function in the previous figure is subtracted from L (we subtract the graph of the exponential from the line P = L).

The equation dP / dt = k ( L - P ), where dP / dt is a first derivative, is an example of a first-order differential equation.

• This first-order equation has a solution which contains one integration constant C.
• When this integration constant is evaluated for a given situation, we obtain a specific function P(t) modeling the population.
• In general, a first-order differential equation has a single constant to be evaluated, which allows us to impose one condition on the solution (e.g., one initial value).

The second equation in the figure below, m d²x / dt², is called a second-order differential equation because its highest derivative is a second derivative.

• The solution of this equation can be either a sine function or a cosine function, as indicated in the fifth line.
• If we add the sine and cosine functions, we still obtain a solution, as you can verify by substituting the indicated x(t) function into the equation.
• The sine and cosine functions, in their most general form, need not have the same coefficient. So we use coefficients A and B for these two functions.
• We say that this solution has parameters A and B.

• We can use trigonometric identities to express the sum of these functions in terms of the parameters C and t₀.

• In either case, both parameters must be evaluated in order to obtain the most general solution.
• In order to evaluate both parameters, we will need two conditions on the solution.   These conditions often take the form of one condition on the solution function itself, and one condition on the first derivative of the solution function.

The equation dy / dx = -1/2 y tells us that the solution function y(x) has a derivative that is equal to -1/2 y, or -1/2 its value.

• This means that everywhere on the line y = 2, for example, the derivative of the function must be -1/2 * 2 = -1.
• As a result, if a solution curve passes through the line y = 2, it must have slope equal to the derivative -1.

• Similarly, everywhere on the line y = 1 the slope must be equal to the derivative -1/2.

• On each of the horizontal lines indicated below we therefore sketch a short segment indicating the slope of a solution curve passing through that line.

• We proceed to sketch curves which pass through the indicated horizontal lines at the indicated slopes.
• We obtain a family of estimated solution curves for the differential equation.

We can start such a curve at any point in the plane.

• A point in the plane consists of an x value and a corresponding y value.
• Specifying an x and a y value is equivalent to specifying an initial condition.

We can compare our family of curves to the general solution y = C e^{(-1/2 x)} of the differential equation.

We see that the curves have the same behavior as the functions y = C e^{(-1/2 x)}.

We recall the Euler method of solving a differential equation approximately.

• We start at a point, like (1,3) in the figure below.
• We calculate the slope at the point based on the differential equation.
• We follow that slope for some specified distance (in the figure below we follow that slope for 1 unit).

• We arrive at a new point (2, 2 1/2) in the figure below, and repeat the process.

• We continue the process until we arrive at the x value at which we wish to find the approximate value of y.

The smaller the x distance we travel with each approximation, the more accurate will be our y values.