Differential Equations: Introduction and Applications, Lecture notes of Differential Equations

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Differential Equations

D. DeTurck

University of Pennsylvania

February 27, 2018

Differential equations

The most important application of integrals is to the solution of differential equations.

From a formal mathematical point of view, a differential equation is an equation that describes a relationship among a function, its independent variable, and the derivative(s) of the function.

For example: dy dx

= 3xy 2

d^2 y dx^2

dy dx

• 3y = 1

Order = highest derivative: first order, second order...

Example 1

The acceleration of gravity is constant (near the surface of the earth). So, for falling objects:

the rate of change of velocity is constant dv dt =g.

Since velocity is in turn the rate of change of position, we could write this as a second-order equation:

d^2 x dt^2 = g.

Example 2

More realistically: With air resistance the acceleration of a falling object is the acceleration of gravity minus the acceleration due to air resistance. For some objects the air resistance is proportional to the square of the velocity. So for such an object we have the differential equation:

The rate of change of velocity is gravity minus something proportional to velocity squared:

dv dt

=g − kv 2 or d^2 x dt^2

= g − k

dx dt

Other situations that yield the same equation

In the presence of abundant resources (food, space), the organisms in a population will reproduce as fast as they can — this means that The rate of change (increase) of the population is proportional to the population itself: dP dt

=kP.

The balance in an interest-paying bank account increases at a rate (called the interest rate) that is proportional to the current balance. So dB dt

= kB.

More realistic situations for the last two problems

For populations: An ecosystem may have a maximum capacity to support a certain kind of organism (we worry about this very thing for people on the planet!)

In this case, the rate of change of population is proportional both to the number of organisms present and to the amount of unused capacity in the environment (overcrowding will cause the population to decrease).

If the carrying capacity of the environment is the constant Pmax , then we get the equation:

dP dt = kP(Pmax − P).

One more application:

Newton’s law of cooling: According to Newton’s law of cooling, the temperature of a hot or cold object will change at a rate proportional to the difference between the object’s temperature and the ambient temperature

If the ambient temperature is kept constant at A, and the object’s temperature is u(t), what is the differential equation for u(t)?

Solving differential equations

Since the the process of solving of a differential equation recovers a function from knowing something about its derivative, it’s not too surprising that we have to use integrals to solve differential equations.

And since we’re using integrals, we should also expect to see some “arbitrary ” constants in the solutions of differential equations. In general, there will be one constant in the solution of a first-order equation, two in a second-order one, etc... In practice...

... we can solve for the constants by having some information about the value of the unknown function (and/or the value of its derivative(s)) at some point. From an applications point of view, such initial conditions are clearly needed, since you can’t determine the value of something just from information about how it is changing. You also need to know its value at some (“initial”) time.

Solving an initial-value problem

So far, we have solved the differential equation dy dx

= 3xy 2 and

obtained −

y

x^2 + C (We only need one constant of integration). This is called the “general solution” of the differential equation.

We can determine C if we are given one point on the graph of the function y (x). For instance, given that y (1) = 2, we would substitute 2 for y and 1 for x in the general solution and get − 12 = 32 + C and conclude that C = −2, so the solution of the initial-value problem dy dx = 3xy 2 , y (1) = 2 is −

y

x^2 − 2 , or better, y =

4 − 3 x^2

Growth and decay

What is the solution of the differential equation:

y ′^ = ky?

How about the initial-value problem:

y ′^ = ky , y (0) = y 0?

As noted previously, this differential equation (of exponential growth and decay ) is useful for modeling radioactive decay, compound interest and unrestricted population growth.

Differential equations greatest hits,

The salty water in the tank problem! A tank contains 1000 liters of brine (salty water) with 50kg of dissolved salt. Pure water enters the tank at the rate of 25 liters per minute. The solution is kept thoroughly mixed and drains at the same rate. How many kg of salt remain in the tank after 10 minutes? The first step in most DiffEq problems is to identify the unknown function. Since we want to know the amount of salt at various times, we’ll write A(t) for the amount of salt (in kg) in the tank at time t (minutes).

We are given that A(0) = 50, which provides the initial condition.

Now we have to produce the differential equation...

Mixture problem differential equation

The rate of change of A(t) could come from salt being added to the tank (but there is none), or from salt flowing out of the tank. The solution flows out at 25 liters per minute and since there are A(t) kg of salt in all 1000 liters of brine, there are A(t)/40 kg in 25 liters.

So, which of the following is the differential equation for this problem?

A. A′^ = A/ 40 B. A′^ = A − 40 C. A′^ = 40 − A D. A′^ = −A/ 40 E. A′^ = − 40 /A

One more greatest hits problem:

For obvious reasons, the dissecting room of a medical examiner is kept very cool, at a constant temperature of 5 degrees C. While doing an autopsy early one morning, the medical examiner himself is killed. At 10 am, the examiner’s assistant discovers the body and finds its temperature to be 23 degrees C, and at noon the body’s temperature is down to 18.5 degrees C. Assuming that the medical examiner had a normal temperature of 37 degrees C when he was alive, when was he murdered?

A. 3 am B. 4 am C. 5 am D. 6 am E. 7 am F. 8 am G. 9 am

Linear differential equations

The other type of DE that we will be able to solve are linear equations. They are of the form:

y ′^ + P(x)y = Q(x) It is important that the coefficient of the y ′^ term is 1. The answer The solution of the equation

y ′^ + P(x)y = Q(x)

is y = e−^

´ P^ (ˆ

e

´ (^) P Q + C

where

P is any anti-derivative of P.

Why does this work? Because e

´ P (^) y ′ (^) + Py =

e

´ P (^) y