Differential Equations
Objective: To solve a separable
differential equation.
Differential Equations
• We will now consider another way of looking at
integration. Suppose that f(x) is a known function
and we are interested in finding a function F(x) such
that y = F(x) satisfies the equation dy  f (x)
dx
Differential Equations
• We will now consider another way of looking at
integration. Suppose that f(x) is a known function
and we are interested in finding a function F(x) such
that y = F(x) satisfies the equation dy  f (x)
dx
• The solutions of this equation are the antiderivatives
of f(x), and we know that these can be obtained by
integrating f(x). For example, the solutions of the
equation dy  x 2 are
x3
2
dx
y   x dx 
3
C
Differential Equations
dy
 f (x )
dx
• An equation of the form
is called a
differential equation because it involves a derivative
of an unknown function. Differential equations are
different from kinds of equations we have
encountered so far in that the unknown is a function
and not a number.
Differential Equations
• Sometimes we will not be interested in finding all of
the solutions of the equation, but rather we will want
only the solution whose integral curve passes through
a specified point.
Differential Equations
• Sometimes we will not be interested in finding all of
the solutions of the equation, but rather we will want
only the solution whose integral curve passes through
a specified point.
• For simplicity, it is common in the study of differential
dy
equations to denote a solution of dx  f (x) as y(x)
rather than F(x), as earlier. With this notation, the
problem of finding a function y(x) whose derivative is
f(x) and whose integral curve passes through the point
(x0, y0) is expressed as dy
dx
 f ( x), y ( x0 )  y0
Differential Equations
dy
• Equations of the form  f ( x), y( x0 )  y0 are known as
dx
initial value problems, and y( x0 )  y0 is called the
initial condition for the problem.
Differential Equations
dy
• Equations of the form  f ( x), y( x0 )  y0 are known as
dx
initial value problems, and y( x0 )  y0 is called the
initial condition for the problem.
• To solve an equation of this type, first we will
separate the variables, integrate, and solve for C.
Example
• Solve the initial-value problem
dy
 cos x, y (0)  1
dx
Example
• Solve the initial-value problem
• Separate the variables
dy
 cos x, y (0)  1
dx
dy  cos xdx
Example
• Solve the initial-value problem
• Separate the variables
• Integrate
dy
 cos x, y (0)  1
dx
dy  cos xdx
 dy   cos xdx
y  sin x  C
Example
• Solve the initial-value problem
• Separate the variables
dy
 cos x, y (0)  1
dx
dy  cos xdx
• Integrate
 dy   cos xdx
y  sin x  C
• Solve for C
1  sin( 0)  C
y  sin x  1
C 1
Example 1
• Solve the initial-value problem dy  4 xy2 , y (0)  1
dx
Example 1
• Solve the initial-value problem dy  4 xy2 , y (0)  1
dx
• Separate the variables
dy
 4 xdx
2
y
Example 1
• Solve the initial-value problem dy  4 xy2 , y (0)  1
dx
• Separate the variables
• Integrate
y
2
dy
 4 xdx
2
y
dy    4 xdx
1
  2 x 2  C
y
1
y 2
2x  C
Example 1
• Solve the initial-value problem dy  4 xy2 , y (0)  1
dx
• Separate the variables
• Integrate
• Solve for C
y
2
dy
 4 xdx
2
y
dy    4 xdx
1
1
0C
C 1
1
  2 x 2  C
y
1
y 2
2x  C
y
1
2x2 1
Example 1
• Solve the initial-value problem dy  4 xy2 , y (0)  1
dx
• Separate the variables
• Integrate
• Solve for C
y
2
dy
 4 xdx
2
y
dy    4 xdx
1
1
0C
C  1
1
  2 x 2  C
y
1
y 2
2x  C
y
1
2x2 1
Example 2
• Solve the initial-value problem
dy
(4 y  cos y )  3 x 2  0, y (0)  0
dx
Example 2
• Solve the initial-value problem
dy
(4 y  cos y )  3 x 2  0, y (0)  0
dx
• Separate the variables
(4 y  cos y)dy  3x 2 dx
Example 2
• Solve the initial-value problem
dy
(4 y  cos y )  3 x 2  0, y (0)  0
dx
• Separate the variables
(4 y  cos y)dy  3x 2 dx
• Integrate
2
(
4
y

cos
y
)
dy

3
x

 dx
2 y 2  sin y  x 3  C
Example 2
• Solve the initial-value problem
dy
(4 y  cos y )  3 x 2  0, y (0)  0
dx
• Separate the variables
(4 y  cos y)dy  3x 2 dx
• Integrate
2
(
4
y

cos
y
)
dy

3
x

 dx
2 y 2  sin y  x 3  C
• Solve for C
2(0) 2  sin(0)  03  C
C 0
x  3 2 y 2  sin y
Example 3
• Find a curve in the xy-plane that passes through (0,3)
and whose tangent line at a point has slope 2 x 2 .
y
Example 3
• Find a curve in the xy-plane that passes through (0,3)
and whose tangent line at a point has slope 2 x 2 .
y
• Since the slope of the tangent line is dy/dx, we have
dy 2 x
 2
dx y
Example 3
• Find a curve in the xy-plane that passes through (0,3)
and whose tangent line at a point has slope 2 x 2 .
y
• Since the slope of the tangent line is dy/dx, we have
 y dy   2 xdx
2
y3
3
 x C
33
3
 02  C
2
C 9
dy 2 x
 2
dx y
Example 3
• Find a curve in the xy-plane that passes through (0,3)
and whose tangent line at a point has slope 2 x 2 .
y
• Since the slope of the tangent line is dy/dx, we have
 y dy   2 xdx
2
y3
3
 x C
3
 02  C
3
3
2
C 9
dy 2 x
 2
dx y
y3
3
 x 9
2
or
y  3 3 x 2  27
Example
• Solve the differential equation
dy
  xy
dx
Example
• Solve the differential equation
• Separate the variables
dy
  xy
dx
dy
  xdx
y
Example
• Solve the differential equation
• Separate the variables
• Integrate
dy
 y    xdx
dy
  xy
dx
dy
  xdx
y
x2
ln | y |   C
2
Example
• Solve the differential equation
dy
  xdx
y
• Separate the variables
• Integrate
• Solve for y
dy
 y    xdx
ye
dy
  xy
dx
x2
 C
2
x2
ln | y |   C
2
Example
• Solve the differential equation
dy
  xdx
y
• Separate the variables
• Integrate
• Solve for y
dy
 y    xdx
ye
dy
  xy
dx
x2
 C
2
x2
ln | y |   C
2
ye
x2

2
 eC  Ce
x2

2
Example 4
• Verify that the given functions are solutions of the
differential equation by substituting the functions
into the equation.
a )e
2 x
b )e x
c)c1e  2 x  c2 e x
y //  y /  2 y  0
Example 4
• Verify that the given functions are solutions of the
differential equation by substituting the functions
into the equation.
a )e
b )e
2 x
x
c)c1e
4e
2 x
y  e 2 x
y  y  2y  0
//
 c2 e
x
/
2 x
 2e
2 x
 2e
2 x
0
y /  2 e  2 x
y //  4e  2 x
Example 4
• Verify that the given functions are solutions of the
differential equation by substituting the functions
into the equation.
a )e
b )e
2 x
y  y  2y  0
y  ex
x
e  e  2e  0
y/  ex
c)c1e
//
x
2 x
 c2 e
x
/
x
x
y //  e x
Example 4
• Verify that the given functions are solutions of the
differential equation by substituting the functions
into the equation.
a )e
b )e
2 x
y  c1e 2 x  c2 e x
x
y /  2c1e  2 x  c2 e x
c)c1e
2 x
 c2 e
x
y //  4c1e  2 x  c2 e x
y //  y /  2 y  0
4c1e 2 x  c2e x  2c1e 2 x  c2e x  2c1e 2 x  2c2e x  0
Exponential Growth and Decay
• Population growth is an example of a general class of
models called exponential models. In general,
exponential models arise in situations where a
quantity increases or decreases at a rate that is
proportional to the amount of the quantity present.
This leads to the following definition:
Exponential Growth and Decay
Exponential Growth and Decay
• Equations 10 and 11 are separable since they have
the right form, but with t rather than x as the
independent variable. To illustrate how these
equations can be solved, suppose that a positive
quantity y = y(t) has an exponential growth model
and that we know the amount of the quantity at
some point in time, say y = y0 when t = 0. Thus, a
formula for y(t) can be obtained by solving the initialvalue problem
dy
 ky, y (0)  y0
dt
Exponential Growth and Decay
• Equations 10 and 11 are separable since they have
the right form, but with t rather than x as the
independent variable. To illustrate how these
equations can be solved, suppose that a positive
quantity y = y(t) has an exponential growth model
and that we know the amount of the quantity at
some point in time, say y = y0 when t = 0. Thus, a
formula for y(t) can be obtained by solving the initialvalue problem
dy
 ky, y (0)  y0
dt

1
y
dy   kdt
ln y  kt  C
Exponential Growth and Decay
• The initial condition implies that y = y0 when t =0.
Solving for C, we get
ln y  kt  C
ln y0  k (0)  C
C  ln y0
Exponential Growth and Decay
• The initial condition implies that y = y0 when t =0.
Solving for C, we get
ln y  kt  C
ln y  kt  ln y0
ln y0  k (0)  C
ye
C  ln y0
y  e e
kt  ln y0
kt
y  y0 e kt
ln y0
Exponential Growth and Decay
• The significance of the constant k in the formulas can
be understood by reexamining the differential
equations that gave rise to these formulas. For
example, in the case of the exponential growth dy
model, we can rewrite the equation as
k  dt
y
which states that the growth rate as a fraction of the
entire population remains constant over time, and
this constant is k. For this reason, k is called the
relative growth rate.
Example 4
• According to United Nations data, the world
population in 1998 was approximately 5.9 billion and
growing at a rate of about 1.33% per year. Assuming
an exponential growth model, estimate the world
population at the beginning of the year 2023.
Example 4
• According to United Nations data, the world
population in 1998 was approximately 5.9 billion and
growing at a rate of about 1.33% per year. Assuming
an exponential growth model, estimate the world
population at the beginning of the year 2023.
y  y0 e kt
y  5.9e (.0133)( 25)
y  8.2billion
Doubling and Half-Life
Doubling Time and Half-Life
• If a quantity has an exponential growth model, then
the time required for the original size to double is
called the doubling time, and the time required to
reduce by half is called the half-life. As it turns out,
doubling time and half-life depend only on the
growth or decay rate and not on the amount present
kT
1
initially.
y

y
e
kT
0
2 0
2 y0  y0 e
2  e kT
ln 2  kT
T  1k ln 2
1
2
 e kT
ln 12  kT
T  1k ln 12
T   1k ln 2
Example 5
• It follows that from the equation that with a
continued growth rate of 1.33% per year, the
doubling time for the world population will be
T  1k ln 2
1
T  .0133
ln 2
T  52.116
Example 5
• It follows that from the equation that with a
continued growth rate of 1.33% per year, the
doubling time for the world population will be
T  1k ln 2
1
T  .0133
ln 2
T  52.116
• If it was a decay rate of 1.33%, we would do the
same thing to find the half-life.
Radioactive Decay
• It is a fact of physics that radioactive elements
disintegrate spontaneously in a process called
radioactive decay. Experimentation has shown that
the rate of disintegration is proportional to the
amount of the element present, which implies that
the amount y = y(t) of a radioactive element present
as a function of time has an exponential decay
model.
Radioactive Decay
• It is a fact of physics that radioactive elements
disintegrate spontaneously in a process called
radioactive decay. Experimentation has shown that
the rate of disintegration is proportional to the
amount of the element present, which implies that
the amount y = y(t) of a radioactive element present
as a function of time has an exponential decay
model. The half life of carbon-14 is 5730 years. The
rate of decay is: T   1 ln 2
k
k   lnT2
k
ln 2
5730
 .000121
Example 6
• If 100 grams of radioactive carbon-14 are stored in a
cave for 1000 years, how many grams will be left at
that time?
Example 6
• If 100 grams of radioactive carbon-14 are stored in a
cave for 1000 years, how many grams will be left at
that time?
y  y0 e
kt
y  100e
( .000121)(1000 )
y  88.6 grams
Carbon Dating
• When the nitrogen in the Earth’s upper atmosphere
is bombarded by cosmic radiation, the radioactive
element carbon-14 is produced. This carbon-14
combines with oxygen to form carbon dioxide, which
is ingested by plants, which in turn are eaten by
animals. In this way all living plants and animals
absorb quantities of radioactive carbon-14.
Carbon Dating
• In 1947 the American nuclear scientist W. F. Libby
proposed the theory that the percentage of carbon14 in the atmosphere and in living tissues of plants is
the same. When a plant or animal dies, the carbon14 in the tissue begins to decay. Thus, the age of an
artifact that contains plant or animal material can be
estimated by determining what percentage of its
original carbon-14 remains. This is called carbondating or carbon-14 dating.
Example 7
• In 1988 the Vatican authorized the
British Museum to date a cloth relic
know as the Shroud of Turin, possibly
the burial shroud of Steve of Nazareth.
This cloth, which first surfaced in 1356,
contains the negative image of a
human body that was widely believed
to be that of Jesus.
Example 7
• The report of the British Museum showed that the
fibers in the cloth contained approximately 92% of
the original carbon-14. Use this information to
estimate the age of the shroud.
Example 7
• The report of the British Museum showed that the
fibers in the cloth contained approximately 92% of
the original carbon-14. Use this information to
estimate the age of the shroud.
y  y0 e kt
y
y0
 e kt
   kt
ln  
t
ln
y
y0
y
y0
k
ln .92 
 689
 .000121
Homework
•
•
•
•
Page 567 (section 8.1)
11
Page 575-576 (section 8.2)
1, 3, 11, 13, 29,31