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Section 15.3 Separable Differential Equations
D1.3 Separable Differential Equations
Sketching solutions of a differential equation using its direction field is a powerful technique, and it provides a wealth of
information about the solutions. However, valuable as they are, direction fields do not produce the actual solutions of a
differential equation. In this section, we examine methods that lead to the solutions of certain differential equations in terms
of an algebraic expression (often called an analytical solution). The equations we consider are first order and belong to a
class called separable equations.
Method of Solution
The most general first-order differential equation has the form y ' (t) = f (t, y), where f (t, y) is an expression that may involve
both the independent variable t and the unknown function y. We have a chance of solving such an equation if it can be
written in the form
g ( y ) y ' (t ) = h (t ).
In the equation g( y) y ' (t) = h(t), the factor g( y) involves only y, and h(t) involves only t; that is, the variables have been
separated. An equation that can be written in this form is said to be separable.
Note
If the equation has the form y ' (t) = f (t) (that is, the right side depends only on t), then
solving the equation amounts to finding the antiderivatives of f .
In general, we solve a separable differential equation by integrating both sides of the equation with respect to t:
 g ( y ) y ' (t ) d t =  h (t ) d t
dy
Integrate both sides with respect to t.
 g( y) d y =  h(t) d t. Change variables on left; d y = y ' (t) d t.
The fact that d y = y ' (t) d t on the left side of the equation leaves us with two integrals to evaluate, one with respect to y and
one with respect to t. Finding a solution depends on evaluating these integrals.
Which of the following equations are separable? (A) y ' (t) = y + t, (B) y ' (t) =
QUICK CHECK 1
y ' ( x) = e x+y
✓
EXAMPLE 1
ty
t+1
, and (C)
A separable equation
Find a function that satisfies the following initial value problem.
1
y ' (t) = y2 e-t , y(0) = , for t ≥ 0.
2
SOLUTION
The equation is written in separable form by dividing both sides of the equation by y2 to give
both sides of the equation with respect to t and evaluate the resulting integrals:
Note
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y ' (t )
y2
= e-t . We now integrate
1
2
Chapter 15 • Differential Equations
In practice, the change of variable on the left side is often omitted, and we go directly to
the second step, which is to integrate the left side with respect to y and the right side with
respect to t.

1
y2
y ' (t) d t =  e-t d t
dy
y2

Note
1
y
=  e-t d t
Change variables on left side.
= -e-t + C. Evaluate integrals.
Notice that each integration produces a constant of integration. The two constants of
integration may be combined into one.
Solving for y gives the general solution
y (t ) =
The initial condition y(0) =
1
2
1
e-t - C
.
implies that
y (0 ) =
1
e0 - C
=
1
1-C
It follows that C = -1, so the solution to the initial value problem is
y (t ) =
The solution (Figure D1.16) passes through 0,
1
2
1
e-t + 1
=
1
2
.
.
and increases to approach the asymptote y = 1 because lim
1
t∞ e-t + 1
= 1.
Figure D1.16
Related Exercises 5–26
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Section 15.3 Separable Differential Equations
QUICK CHECK 2
✓
EXAMPLE 2
Write y ' (t) =
t2 + 1
y3
in separated form.
Another separable equation
Find the solutions of the equation y ' ( x) = e-y sin x subject to the three different initial conditions
y(0) = 1, y
π
2
=
1
2
, and y(0) = -3.
SOLUTION
Writing the equation in the form e y y ' ( x) = sin x, we see that it is separable. Integrating both sides with respect to x, we have
y
 e y ' ( x) d x =  sin x d x
y
 e d y =  sin x d x
Change variables on left side.
e y = -cos x + C. Evaluate integrals.
The general solution y is found by taking logarithms of both sides of this equation:
y = ln (C - cos x).
The three initial conditions are now used to evaluate the constant C for the three solutions:
y (0) = 1  1 = ln(C - cos 0) = ln(C - 1) 
e = C - 1  C = e + 1,
π
1
1
π
y
=

= ln C - cos
= ln C

C = e1/2 , and
2
2
2
2
y(0) = -3  -3 = ln(C - cos 0) = ln(C - 1)  e-3 = C - 1  C = e-3 + 1.
Substituting these values of C into the general solution gives the solutions of the three initial value problems (Figure D1.17).
The points indicate the initial condition for each solution.
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4
Chapter 15 • Differential Equations
C
y
C=e+1
1 (0, 1)
C = e1/2
C = e-3 + 1
C = 1.6487
-2
π 1
,
2 2
2
4
6
8
10
x
-1
-2
-3 (0, -3)
Figure D1.17
Related Exercises 5–26
Find the value of the constant C in Example 2 with the initial condition y(π) = 0.
QUICK CHECK 3
✓
Even if we can evaluate the integrals necessary to solve a separable equation, the solution may not be easily expressed
in an explicit form. Here is an example of a solution that is best left in implicit form.
EXAMPLE 3
An implicit solution
Find and graph the solution of the initial value problem
cos y y ' (t) = sin2 t cos t, y(0) = .
6
π
SOLUTION
The equation is already in separated form. Integrating both sides with respect to t, we have
Note
For the integral on the right side, we use the substitution u = sin t. The integral becomes
1
2
3
 u d u = u + C.
3
2
 cos y d y =  sin t cos t d t Integrate both sides.
sin y =
1
3
sin3 t + C.
Evaluate integrals.
When imposing the initial condition in this case, it is best to leave the general solution in implicit form. Substituting t = 0 and
y=
π
6
into the general solution, we find that
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Section 15.3 Separable Differential Equations
sin
π
6
=
Therefore, the solution of the initial value problem is
5
1
1
sin3 0 + C or C = .
3
2
sin y =
1
1
sin3 t + .
3
2
In order to graph the solution in this implicit form, it is easiest to use graphing software. The result is shown in Figure D1.18.
initial condition
y (0 ) =
5π
y (0 ) =
y (0 ) = -
6
C=
1
y
4
2
0,
π
6
7π
6
5π
6
2
0,
-4
π
6
-2
2
-2
0, -4
7π
6
Figure D1.18
Note
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4 x
6
Chapter 15 • Differential Equations
Care must be used when graphing and interpreting implicit solutions. The graph of
sin y =
1
3
sin3 t +
is a family of an infinite number of curves.
1
2
y
3
2
1
-4 -3 -2 -1
-1
1
2
3
x
-2
-3
-4
You must choose the curve that satisfies the initial condition, as shown in Figure D1.18.
Find the value of the constant C in Example 3 with the initial condition y
QUICK CHECK 4
Related Exercises 27–32
π
6
= 0.
✓
Logistic Equation Revisited
In Section D1.1, we introduced the logistic equation, which is commonly used for modeling populations, epidemics, and the
spread of rumors. In Section D1.2, we investigated the direction field associated with the logistic equation. It turns out that
the logistic equation is a separable equation, so we now have the tools needed to solve it.
Note
The derivation of the logistic equation is discussed in Section D1.5.
EXAMPLE 4
Logistic population growth
Assume 50 fruit flies are in a large jar at the beginning of an experiment. Let P(t) be the number of fruit flies in the jar t days
later. At first, the population grows exponentially, but due to limited space and food supply, the growth rate decreases and
the population is prevented from growing without bound. This experiment is modeled by the logistic equation
dP
dt
= 0.1 P 1 -
P
300
, for t ≥ 0,
together with the initial condition P(0) = 50. Solve this initial value problem.
SOLUTION
We see that the equation is separable by writing it in the form
1
P
P 1 
300
·
dP
dt
= 0.1.
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7
Integrating both sides with respect to t leads to the equation
dP

P

300
P 1 -
=  0.1 d t.
(1)
The integral on the right side of equation (1) is  0.1 d t = 0.1 t + C.
Because the integrand on the left side is a rational function in P, we use partial fractions. You should verify that
1
P 1 and therefore,

1
P 1 -
After integration, equation (1) becomes
P

300
P

300
=
dP = 
ln
300
P (300 - P)
1
P
1
+
=
300 - P
P
300 - P
1
P
+
1
300 - P
d P = ln
P
300 - P
+ C.
= 0.1 t + C.
(2)
The next step is to solve for P, which is tangled up inside the logarithm. We first exponentiate both sides of equation (2) to
obtain
P
300 - P
= eC · e0.1 t .
We can remove the absolute value on the left side of equation (2) by writing
P
300 - P
= ± eC · e0.1 t .
At this point, a useful trick simplifies matters. Because C is an arbitrary constant, ± eC is also an arbitrary constant, we
rename ± eC as C. We now have
P
300 - P
Note
= C e0.1 t .
(3)
There are not many times in mathematics when we can redefine a constant in the middle
of a calculation. When working with arbitrary constants, it may be possible, if it is done
carefully.
Solving equation (3) for P and replacing
1
C
by C gives the general solution
P (t ) =
300
1 + C e-0.1 t
.
Figure D1.19 shows the general solution, with curves corresponding to several different values of C. Using the initial
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Figure D1.19 shows the general solution, with curves corresponding to several different values of C. Using the initial
condition P(0) = 50, we find the value of C for our specific problem is C = 5. It follows that the solution of the initial value
problem is
P (t ) =
Note
300
1 + 5 e-0.1 t
.
We could also use the initial condition in equation (3) to solve for C.
C
P
P (t ) =
300
C = -8
300
1 - 8 e-0.1 t
200
P (t ) =
100
300
1 + 5 e-0.1 t
20
40
60
80
t
Figure D1.19
Figure D1.19 also shows this particular solution (in red) among the curves in the general solution. A significant feature
of this model is that, for 0 < P(0) < 300, the population increases, but not without bound. Instead, it approaches an equilibrium, or steady-state, solution with a value of
lim P(t) = lim
t∞
t∞
300
1 + 5 e-0.1 t
= 300,
which is the maximum population that the environment (space and food supply) can sustain. This equilibrium population is
called the carrying capacity. Notice that all the curves in the general solution approach the carrying capacity as t increases.
Related Exercises 33–34
Quick Quiz
1. Which of the following is a separable differential equation?
(a) y + y' = t
(b) y · y' = y + t
(c) y · y' = t csc y
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2. A differential equation of the form g(t) y' (t) = h(t) is
(a) first order and separable.
(b) second order and separable.
(c) non-separable.
3. Which of the following familes of functions satisfies the equation y ' = 2 t y?
2
(a) y = et + C
(b) y =
1
2
t2 y2 + C
(c) y = C et
2
4. The solution of the initial value problem y ' = t e- y , y(0) = 1 is
t 2 2
(a) y = e
.
(b) y = ln
1
2
t2 + e .
1
(c) y = - ln e -
2
t2 .
5. The general solution of the equation k y' + b = 0, for k ≠ 0, is a family of curves, all of which are
(a) parabolas.
(b) exponential curves.
(c) lines.
6. The solution of the initial value problem y ' - y-3 = 0, y(1) = 2 is
(a) y =
4
1
(b) y =
(c) y =
4 t + 12 .
4
3
t+
4t +
11
4
3
.
12 .
7. The general solution of y ' = 2 t
1
y is
t2 + C.
4
2
1
(b)
t2 + C .
2
1
(c)
C t2 .
2
(a)
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8. An implicit solution of y ' =
(a) y4 + y3 = tan t + 1.
sec2 t
4 y3 + 3 y2
,y
π
4
= 1 is
(b) y4 + y3 - tan t = 0.
(c) y = 4 tan t - y3 .
SECTION D1.3 EXERCISES
Review Questions
1.
What is a separable first-order differential equation?
2.
Is the equation t2 y ' (t) =
3.
Is the equation y ' (t) = 2 y - t separable?
4.
Explain how to solve a separable differential equation of the form g(t) y ' (t) = h(t).
t+4
y2
separable?
Basic Skills
5–16. Solving separable equations Find the general solution of the following equations. Express the solution
explicitly as a function of the independent variable.
5.
t-3 y ' (t) = 1
6.
e 4 t y ' (t ) = 5
7.
8.
9.
dy
dt
dy
dx
=
3 t2
y
= y  x 2 + 1
y ' (t) = e y/2 sin t
10. x2
dw
dx
=
w (3 x + 1 ), x > 0
11. x2 y ' ( x) = y2 , x > 0
12. t2 + 13 y y ' (t) = t  y2 + 4
13. y ' (t) csc t =
- y3
2
14. y ' (t) et/2 = y2 + 4
15. u ' ( x) = e2 x-u
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16. x u ' ( x) = u2 - 4, x > 0
17–26. Solving initial value problems Determine whether the following equations are separable. If so, solve the
initial value problem.
17. t y ' (t) = 1, y(1) = 2, t > 0
18. sec t y ' (t) = 1, y(0) = 1
19. 2 y y ' (t) = 3 t2 , y(0) = 9
20. y ' (t) = et y , y(0) = 1
21.
dy
dt
= t y + 2, y(1) = 2
22. y ' (t) = y 4 t3 + 1, y(0) = 4
23. y ' (t) =
et
2y
, y(ln 2) = 1
24. sec x y ' ( x) = y3 , y(0) = 3
25.
dy
dx
= e x-y , y(0) = ln 3
26. y ' (t) = cos2 y, y(1) =
T
π
4
27-32. Solutions in implicit form Solve the following initial value problems and leave the solution in implicit form.
Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate
which curve corresponds to the solution of the initial value problem.
27. y ' (t) =
28. y ' ( x) =
t
y
, y (1 ) = 2
1+x
2-y
, y (1 ) = 1
x
π
29. u ' ( x) = csc u cos , u(π) =
2
2
30. y y ' ( x) =
31. y ' ( x) =
32. z' ( x) =
2x
2 + y2 2
x+1
y+4
z2 + 4
x2 + 16
, y(1) = -1
, y (3 ) = 5
, z(4) = 2
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Chapter 15 • Differential Equations
T
33. Logistic equation for a population A community of hares on an island has a population of 50 when
observations begin (at t = 0). The population is modeled by the initial value problem
dP
P
= 0.08 P 1 , P(0) = 50.
dt
200
a. Find and graph the solution of the initial value problem, for t ≥ 0.
b. What is the steady-state population?
T
34. Logistic equation for an epidemic When an infected person is introduced into a closed and otherwise healthy
community, the number of people who contract the disease (in the absence of any intervention) may be modeled
by the logistic equation
dP
P
=kP 1, P (0 ) = P 0 ,
dt
A
where k is a positive infection rate, A is the number of people in the community, and P0 is the number of infected
people at t = 0. The model also assumes no recovery.
a. Find the solution of the initial value problem, for t ≥ 0, in terms of k, A, and P0 .
b. Graph the solution in the case that k = 0.025, A = 300, and P0 = 1.
c. For a fixed value of k and A, describe the long-term behavior of the solutions, for any P0 with 0 < P0 < A.
Further Explorations
35. Explain why or why not Determine whether the following statements are true and give an explanation or
counterexample.
a. The equation u ' ( x) =  x2 u7 -1 is separable.
b. The general solution of the separable equation y ' (t) =
t
y7 + 10 y4
can be expressed explicitly with y in terms
of t.
c. The general solution of the equation y y ' ( x) = x e-y can be found using integration by parts.
36–39. Solutions of separable equations Solve the following initial value problems. When possible, give the
solution as an explicit function of t.
36. e y y ' (t) =
37. y ' (t) =
38. y ' (t) =
39. y ' (t) =
ln2 t
t
, y(1) = ln 2
3 y ( y + 1)
t
cos2 t
2y
y+3
, y (1 ) = 1
, y(0) = -2
5t+6
, y (2 ) = 0
40–41. Implicit solutions for separable equations For the following separable equations, carry out the indicated
analysis.
a. Find the general solution of the equation.
b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition
requires a different constant.)
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c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition.
40. y2 y ' (t) = t2 +
2
3
t; y(-1) = 1, y(1) = 0, y(-1) = -1
41. e-y/2 y ' ( x) = 4 x sin x2 - x; y(0) = 0, y(0) = ln
1
4
,y
π
2
=0
42. Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at
each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if
each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the
orthogonal trajectories of the family of ellipses 2 x2 + y2 = a2 .
a. Apply implicit differentiation to 2 x2 + y2 = a2 to show that
dy
dx
=
-2 x
y
.
b. The family of trajectories orthogonal to 2 x2 + y2 = a2 satisfies the differential equation
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dy
dx
=
y
2x
. Why?
13
14
Chapter 15 • Differential Equations
c.
Solve the differential equation in part (b) to verify that y2 = eC  x and then explain why it follows that
y2 = k x. Therefore, the family of parabolas y2 = k x forms the orthogonal trajectories of the family of
ellipses 2 x2 + y2 = a2 .
43. Orthogonal trajectories Use the method in Exercise 42 to find the orthogonal trajectories for the family of
circles x2 + y2 = a2 .
Applications
T
44. Logistic equation for spread of rumors Sociologists model the spread of rumors using logistic equations. The
key assumption is that at any given time, a fraction y of the population, where 0 ≤ y ≤ 1, knows the rumor, while
the remaining fraction 1 - y does not. Furthermore, the rumor spreads by interactions between those who know
the rumor and those who do not. The number of such interactions is proportional to y (1 - y). Therefore, the
equation that describes the spread of the rumor is y ' (t) = k y (1 - y), where k is a positive real number. The
number of people who initially know the rumor is y(0) = y0 , where 0 ≤ y0 ≤ 1.
a. Solve this initial value problem and give the solution in terms of k and y0 .
b. Assume k = 0.3 weeks-1 and graph the solution for y0 = 0.1 and y0 = 0.7.
c. Describe and interpret the long-term behavior of the rumor function, for any 0 ≤ y0 ≤ 1.
T
45. Free fall An object in free fall may be modeled by assuming that the only forces at work are the gravitational
force and air resistance. By Newton's Second Law of Motion (mass × acceleration = the sum of external forces),
the velocity of the object satisfies the differential equation
m · v ' (t ) = m g + f (v ) ,

mass acceleration
external forces
where f is a function that models the air resistance (assuming the positive direction is downward). One common
assumption (often used for motion in air) is that f (v) = -k v2 , where k > 0 is a drag coefficient.
a. Show that the equation can be written in the form v ' (t) = g - a v2 , where a =
k
.
m
b. For what (positive) value of v is v ' (t) = 0? (This equilibrium solution is called the terminal velocity.)
g
c. Find the solution of this separable equation assuming v(0) = 0 and 0 < v2 < .
a
2
d. Graph the solution found in part (c) with g = 9.8 m  s , m = 1, and k = 0.1, and verify that the terminal
velocity agrees with the value found in part (b).
T
46. Free fall Using the background given in Exercise 45, assume the resistance is given by f (v) = -R v, where R > 0
is a drag coefficient (an assumption often made for a heavy medium such as water or oil).
a. Show that the equation can be written in the form v ' (t) = g - b v, where b =
R
.
m
b. For what value of v 9s v ' (t) = 0? (This equilibrium solution is called the terminal velocity.)
g
c. Find the solution of this separable equation assuming v(0) = 0 and 0 < v < .
b
d. Graph the solution found in part (c) with g = 9.8 m  s2 , m = 1, and R = 0.1, and verify that the terminal
velocity agrees with the value found in part (b).
T
47. Torricelli's law An open cylindrical tank initially filled with water drains through a hole in the bottom of the
tank according to Torricelli's Law (see figure). If h(t) is the depth of water in the tank, for t ≥ 0, then Torricelli's
Law implies h ' (t) = -2 k
h , where k is a constant that includes g = 9.8 m  s2 , the radius of the tank, and the
radius of the drain. Assume that the initial depth of the water is h(0) = H.
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a. Find the solution of the initial value problem.
b. Find the solution in the case that k = 0.1 and H = 0.5 m.
c. In part (b), how long does it take for the tank to drain?
d. Graph the solution in part (b) and check that it is consistent with part (c).
T
48. Chemical rate equations Let y(t) be the concentration of a substance in a chemical reaction (typical units are
moles / liter). The change in the concentration, under appropriate conditions, is modeled by the equation
dy
= -k yn , where k > 0 is a rate constant and the positive integer n is the order of the reaction.
dt
a. Show that for a first-order reaction (n = 1), the concentration obeys an exponential decay law.
b. Solve the initial value problem for a second-order reaction (n = 2) assuming y(0) = y0 .
c. Graph the concentration for a first-order and second-order reaction with k = 0.1 and y0 = 1.
T
49. Tumor growth The Gompertz growth equation is often used to model the growth of tumors. Let M (t) be the
mass of a tumor at time t ≥ 0. The relevant initial value problem is
dM
M
= -r M ln
, M (0 ) = M 0 ,
dt
K
where r and K are positive constants and 0 < M0 < K.
a. Graph the growth rate function R(M ) = -r M ln
M
(which equals M ' (t)) assuming r = 1 and K = 4. For
K
what values of M is the growth rate positive? For what value of M is the growth rate a maximum?
b. Solve the initial value problem and graph the solution for r = 1, K = 4, and M0 = 1. Describe the growth
pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor?
c. In the general solution, what is the meaning of K?
Additional Exercises
T
50. Technology for an initial value problem Solve y ' (t) = y et cos3 4 t, y(0) = 1, and plot the solution for 0 ≤ t ≤ π.
51. Blowup in finite time Consider the initial value problem y ' (t) = yn+1 , y(0) = y0 , where n is a positive integer.
a. Solve the initial value problem with n = 1 and y0 = 1.
b. Solve the initial value problem with n = 2 and y0 =
1
2
.
c. Solve the problem for positive integers n and y0 = n-1/n . How do solutions behave as t  1- ?
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Pearson Student
Education,
15
16
Chapter 15 • Differential Equations
T
52. Analysis of a separable equation Consider the differential equation y ' (t) =
following analysis.
y ( y + 1)
t (t + 2 )
and carry out the
a. Show that the general solution of the equation can be written in the form
y (t ) =
t
C
t+2 -
t
.
b. Now consider the initial condition y(1) = A, where A is a real number. Show that the solution of the initial
value problem is
y (t ) =
1+A
3 A
t
t+2 -
t
.
c. Find and graph the solution that satisfies the initial condition y(1) = 1.
d. Describe the behavior of the solution in part (c) as t increases.
e. Find and graph the solution that satisfies the initial condition y(1) = 2.
f. Describe the behavior of the solution in part (e) as t increases.
g. In the cases in which the solution is bounded for t > 0, what is the value of lim y(t)?
t∞
T
53. Analysis of a separable equation Consider the differential equation y y ' (t) =
following analysis.
1
2
et + t and carry out the
a. Find the general solution of the equation and express it explicitly as a function of t in two cases: y > 0 and
y < 0.
b. Find the solutions that satisfy the initial conditions y(-1) = 1 and y(-1) = 2.
c. Graph the solutions in part (b) and describe their behavior as t increases.
d. Find the solutions that satisfy the initial conditions y(-1) = -1 and y(-1) = -2.
e. Graph the solutions in part (d) and describe their behavior as t increases.
Printed by©
Wolfram
Edition Inc.
Copyright
2014Mathematica
Pearson Student
Education,