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7.1. AN INTRODUCTION TO DIFFERENTIAL EQUATIONS
Solving a differential equation
We have said that a differential equation is an equation that describes the derivative, or
derivatives, of a function that is unknown to us. By a solution to a differential equation, we
mean simply a function that satisies this description.
For instance, the first differential equation we looked at is
ds
dt
= 4t + 1,
which describes an unknown function s(t). We may check that s(t) = 2t 2 + t is a solution
because it satisfies this description. Notice that s(t) = 2t 2 + t + 4 is also a solution.
If we have a candidate for a solution, it is straightforward to check whether it is a
solution or not. Before we demonstrate, however, let’s consider the same issue in a simpler
context. Suppose we are given the equation 2x 2 − 2x = 2x + 6 and asked whether x = 3 is
a solution. To answer this question, we could rewrite the variable x in the equation with
the symbol :
2
2 − 2 = 2 + 6.
To determine whether x = 3 is a solution, we can investigate the value of each side of the
equation separately when the value 3 is placed in and see if indeed the two resulting
values are equal. Doing so, we observe that
2
2 − 2 = 2 · 3
2 − 2 · 3 = 12,
and
2 + 6 = 2 · 3 + 6 = 12.
Therefore, x = 3 is indeed a solution.
We will do the same thing with differential equations. Consider the differential equation
dv
dt
= 1.5 − 0.5v, or
d
dt
= 1.5 − 0.5.
Let’s ask whether v(t) = 3 − 2e −0.5t is a solution 1 . Using this formula for v, observe first
that
dv
dt
=
d
dt
=
d
dt
[3 − 2e
−0.5t ] = −2e
−0.5t · (−0.5) = e
−0.5t
and
1.5 − 0.5v = 1.5 − 0.5 = 1.5 − 0.5(3 − 2e
−0.5t ) = 1.5 − 1.5 + e
−0.5t = e
−0.5t .
1 At this time, don’t worry about why we chose this function; we will learn techniques for finding solutions
to differential equations soon enough.