James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 9.1 Modeling with Differential Equations 589
ever, we will see how to draw rough graphs of solutions even when we have no explicit
formula. We will also learn how to find numerical approximations to solutions.
Example 1 Show that every member of the family of functions
y − 1 1 ce t
1 2 ce t
is a solution of the differential equation y9 − 1 2 sy2 2 1d.
SOLUtion We use the Quotient Rule to differentiate the expression for y:
y9 − s1 2 ce t dsce t d 2 s1 1 ce t ds2ce t d
s1 2 ce t d 2
Figure 5 shows graphs of seven
members of the family in Example 1.
The differential equation shows that if
y < 61, then y9 < 0. That is borne out
by the flatness of the graphs near y − 1
and y − 21.
_5 5
5
− ce t 2 c 2 e 2t 1 ce t 1 c 2 e 2t
s1 2 ce t d 2
The right side of the differential equation becomes
1
2 sy2 2 1d − FS tD
1 1 1 ce t 2
2 1 2 ce
−
2 1G
2ce t
s1 2 ce t d 2
− 1 2
F s1 1 ce t d 2 2 s1 2 ce t d 2
s1 2 ce t d 2
− 1 2
4ce t
s1 2 ce t d 2 − 2ce t
s1 2 ce t d 2
G
FIGURE 5
_5
Therefore, for every value of c, the given function is a solution of the differential
equation.
n
When applying differential equations, we are usually not as interested in finding a
family of solutions (the general solution) as we are in finding a solution that satisfies
some additional requirement. In many physical problems we need to find the particular
solution that satisfies a condition of the form yst 0 d − y 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.
Geometrically, when we impose an initial condition, we look at the family of solution
curves and pick the one that passes through the point st 0 , y 0 d. Physically, this corresponds
to measuring the state of a system at time t 0 and using the solution of the initial-value
problem to predict the future behavior of the system.
Example 2 Find a solution of the differential equation y9 − 1 2 sy2 2 1d that satisfies
the initial condition ys0d − 2.
SOLUtion Substituting the values t − 0 and y − 2 into the formula
from Example 1, we get
y − 1 1 ce t
1 2 ce t
2 − 1 1 ce 0
1 2 ce 0 − 1 1 c
1 2 c
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