Differential Equations Notes
Differential Equations
9.1 Modeling with Differential Equations
A mathematical model often takes the form of a differential equation, which contains an unknown function and some of its derivatives.
Models of Population Growth
A population growth model assumes the population grows at a rate proportional to its size, which is reasonable for bacteria or animals under ideal conditions.
Variables:
t = time (independent variable)
P = number of individuals in the population (dependent variable)
The rate of growth of the population is the derivative \frac{dP}{dt}.
The assumption that the growth rate is proportional to population size can be written as: \frac{dP}{dt} = kP, where k is the proportionality constant.
This is a differential equation because it contains an unknown function P and its derivative \frac{dP}{dt}.
If k > 0, then \frac{dP}{dt} > 0 for all t, meaning the population is always increasing.
As P(t) increases, \frac{dP}{dt} becomes larger, so the growth rate increases as the population increases.
A solution to the equation \frac{dP}{dt} = kP is an exponential function of the form P(t) = Ce^{kt}.
Taking the derivative: P'(t) = C(ke^{kt}) = k(Ce^{kt}) = kP(t)
The family of solutions is P(t) = Ce^{kt}.
Since populations have only positive values, we are interested in solutions with C > 0 and t > 0.
When t = 0, P(0) = Ce^{k(0)} = C, so C is the initial population, P(0).
A more realistic model must account for limited resources, where the population levels off when it approaches its carrying capacity M.
Assumptions for a refined model:
If P is small, the growth rate is proportional to P.
If P > M, then P decreases if it ever exceeds M.
A simple expression incorporating both assumptions is: \frac{dP}{dt} = kP(1 - \frac{P}{M})
If P is small compared with M, then \frac{P}{M} is close to 0, so \frac{dP}{dt} \approx kP.
If P > M, then 1 - \frac{P}{M} is negative, so \frac{dP}{dt} < 0.
The equation \frac{dP}{dt} = kP(1 - \frac{P}{M}) is called the logistic differential equation.
The constant functions P(t) = 0 and P(t) = M are solutions because one of the factors on the right side of the equation is zero; these are equilibrium solutions.
If the initial population P(0) lies between 0 and M, then \frac{dP}{dt} > 0 and the population increases.
If the population exceeds the carrying capacity (P > M), then 1 - \frac{P}{M} is negative, so \frac{dP}{dt} < 0 and the population decreases.
If the population approaches the carrying capacity (P \rightarrow M), then \frac{dP}{dt} \rightarrow 0, which means the population levels off.
A Model for the Motion of a Spring
Consider the motion of an object with mass m at the end of a vertical spring.
Hooke’s Law: If the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x: restoring force = -kx, where k is a positive constant (the spring constant).
Ignoring external resisting forces, by Newton’s Second Law (force equals mass times acceleration), we have m\frac{d^2x}{dt^2} = -kx.
This is a second-order differential equation because it involves second derivatives.
The equation can be rewritten as: \frac{d^2x}{dt^2} = -\frac{k}{m}x, which means the second derivative of x is proportional to x but has the opposite sign.
General Differential Equations
A differential equation is an equation that contains an unknown function and one or more of its derivatives.
The order of a differential equation is the order of the highest derivative that occurs in the equation.
\frac{dP}{dt} = kP and \frac{dP}{dt} = kP(1 - \frac{P}{M}) are first-order equations.
m\frac{d^2x}{dt^2} = -kx is a second-order equation.
The independent variable (e.g., t) doesn’t always have to represent time.
For example, in the differential equation y' = xy, y is an unknown function of x.
A function f is a solution of a differential equation if the equation is satisfied when y = f(x) and its derivatives are substituted into the equation.
f is a solution of y' = xy if f'(x) = xf(x) for all values of x in some interval.
To solve a differential equation means to find all possible solutions of the equation.
We have already solved simple differential equations of the form y' = f(x).
The general solution of y' = x^3 is y = \frac{x^4}{4} + C, where C is an arbitrary constant.
Example 1
Show that every member of the family of functions y = \frac{1 + ce^t}{1 - ce^t} is a solution of the differential equation y' = \frac{1}{2}(y^2 - 1).
Solution
Using the Quotient Rule to differentiate y:
y' = \frac{(1 - ce^t)(ce^t) - (1 + ce^t)(-ce^t)}{(1 - ce^t)^2} = \frac{ce^t - c^2e^{2t} + ce^t + c^2e^{2t}}{(1 - ce^t)^2} = \frac{2ce^t}{(1 - ce^t)^2}The right side of the differential equation becomes:
\frac{1}{2}(y^2 - 1) = \frac{1}{2}[(\frac{1 + ce^t}{1 - ce^t})^2 - 1] = \frac{1}{2}[\frac{(1 + ce^t)^2 - (1 - ce^t)^2}{(1 - ce^t)^2}] = \frac{1}{2}[\frac{1 + 2ce^t + c^2e^{2t} - (1 - 2ce^t + c^2e^{2t})}{(1 - ce^t)^2}] = \frac{1}{2}[\frac{4ce^t}{(1 - ce^t)^2}] = \frac{2ce^t}{(1 - ce^t)^2}Therefore, for every value of c, the given function is a solution of the differential equation.
When applying differential equations, we are usually interested in finding a particular solution that satisfies an additional requirement, rather than a family of solutions (the general solution).
In many physical problems, we need to find the particular solution that satisfies a condition of the form y(t0) = y0.
This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem.