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 $y(t<em>0) = y</em>0$.

  • 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.