Section 9.1 Introduction to Di erential Equations

Section 9.1

Introduction to Differential Equations

A differential equation is an equation involving an unknown function and one or more of its derivatives. For example, the equation

2y = y is an equation involving the unknown function y and its derivative y ; we think of y as a function of some other variable, say x (although t is often used), which we think of as the "controlling variable". Then y is the rate of change of the (unknown) y with respect to x. The equation

2y = xy

is a differential equation as well; it relates the unknown function y, its derivative y , and the controlling variable x.

Since rates of change are inherent in nature, differential equations are used as models for natural phenomena. For instance, the differential equation

dy = ky

dt is used to model the growth of large populations of animals or of bacteria. Differential equations can be used to determine the level of the mixture in a tank when liquids are being added and drained at different rates. They can also be used to determine the proportions of particular chemicals in a mixture based on the rates at which the chemicals are added.

One of the goals of studying differential equations is to be able to solve the types of equations that present themselves in such phenomena. By "solve a differential equation", we mean that we wish to find a specific function y that satisfies the given relationship; the specific funtion y that solves the equation is a solution.

Example. Show that y = x2 is a solution to the differential equation 2y = xy .

Since y = 2x, we see that

xy = x(2x) = 2x2 = 2y.

Example. Show that y = e3x cos 2x is a solution to the differential equation y - 6y + 13y = 0.

Since and

y = 3e3x cos 2x - 2e3x sin 2x

y = 9e3x cos 2x - 6e3x sin 2x - 6e3x sin 2x - 4e3x cos 2x = 5e3x cos 2x - 12e3x sin 2x

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Section 9.1

we have y - 6y + 13y = 5e3x cos 2x - 12e3x sin 2x - 6(3e3x cos 2x - 2e3x sin 2x) + 13(e3x cos 2x) = 5e3x cos 2x - 12e3x sin 2x - 18e3x cos 2x + 12e3x sin 2x + 13e3x cos 2x = 5e3x cos 2x - 18e3x cos 2x + 13e3x cos 2x - 12e3x sin 2x + 12e3x sin 2x = 0.

Example. Find a function y that solves the differential equation 2y = y .

We want to find a function y that is proportional to its derivative y . Thinking through some basic functions, ex comes to mind since

d ex = ex. dx In other words, if z = ex, then z = ex as well, and we have the relationship z = z , which is very similar to the equation we want to solve. It looks like we just need to make a minor alteration: setting y = e2x, we see that y = 2e2x. With this choice of functions, we now have 2y = y as desired, which means that y = e2x is a solution to the differential equation 2y = y .

One minor note here: notice that y = e2x is not the only solution to the equation. For example, y = 5e2x, y = -3e2x, or more generally y = Ce2x, C a constant, would all have y = 2Ce2x, thus would all be solutions to the equation. We call

y = Ce2x the most general solution to the differential equation.

Some problems will ask us to find a specific solution among all of the possible solutions; such a problem is known as an initial value problem.

Example. Find a solution y to the differential equation 2y = y

so that y(0) = 3.

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Section 9.1

We have already determined that the most general solution to the differential equation is y = Ce2x,

so to solve the initial value problem, we need to determine the value for C. Specifying that y(0) = 3 allows us to do this easily: this means that

3 = Ce0 = C,

so C = 3. Thus the solution to the initial value problem is y = 3e2x.

The differential equation 2y = y

can be rewritten as y =

1 2

y

;

we

note

that

that

y

and

y

are proportional, i.e. one is a (constant) multiple of the other. Differential equations of the form

y = ky , k a constant, turn out to be quite important in modeling population change, so it would

be nice to know all possible solutions of such an equation. The following theorem answers the

question:

Theorem. The only solutions of

dy y = ky (which can be rewritten as = ky)

dt

are the exponential functions

y(t) = y(0)ekt.

Example

A population of unicorns increases proportionally to its size. On the second day of the experiment, there were 100 unicorns, and on the fourth day, there were 300 unicorns. Approximately how many unicorns were in the original population?

Let P = P (t) be the number of unicorns at time t. The differential equation that describes the situation is P = kP , whose solution can be written as P (t) = P (0)ekt by the previous theorem. We need to find the value for P (0), and we should be able to use what we are given about the population sizes at t = 2 and t = 4 to do so. We know that

100 = P (2) = P (0)e2k,

and 300 = P (4) = P (0)e4k.

Solving the first equation for P (0), we see that

100 P (0) = e2k ;

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Section 9.1

substituting this in the second equation, we now have

300 =

100 e2k

e4k = 100e2k.

So

e2k

=

300 =3

100

2k = ln 3

ln 3

k =

.

2

Again using the first equation, we see that 100 = P (0)eln 3 = 3P (0),

so that

100 P (0) .

3

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