Download Differential Equations A differential equation is an

Section 9.1
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 ′ . 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.
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 x
e = 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
1
Section 9.1
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 function y so that y(0) = 3 that solves the differential equation
2y = y ′ .
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 .
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
y ′ = 3e3x cos 2x − 2e3x sin 2x
and
y ′′ = 9e3x cos 2x − 6e3x sin 2x − 6e3x sin 2x − 4e3x cos 2x = 5e3x cos 2x − 12e3x sin 2x,
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.
2
Section 9.1
Earlier, we looked at the differential equation 2y = y ′ , which we can rewrite as y = 21 y ′ , and
noted 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 answer turns
out to be quite nice:
Theorem 0.0.1. (6.5 Theorem 2) The only solutions of
y ′ = ky (which can be rewritten as
dy
= 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
P (0) =
100
;
e2k
substituting this in the second equation, we now have
)
(
100 4k
300 = 2k e = 100e2k .
e
So
300
=3
100
⇒ 2k = ln 3
ln 3
.
⇒k =
2
e2k =
3
Section 9.1
Again using the first equation, we see that
100 = P (0)eln 3 = 3P (0),
so that
P (0) ≈
4
100
.
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