Math 308, Sections 301, 302,
Summer 2008
Review before Test I
06/09/2008
Chapter 1. Introduction
Section 1.1 Background
Definition Equation that contains some derivatives of an unknown
function is called a differential equation.
Definition A differential equation involving only ordinary
derivatives with respect to a single variable is called an ordinary
differential equations or ODE. A differential equation involving
partial derivatives with respect to more then one variable is a
partial differential equations or PDE.
Definition The order of a differential equation is the order of the
highest-order derivatives present in equation.
Definition An ODE is linear if it has format
an (x)
d ny
d n−1 y
dy
+ a0 (x)y = F (x),
+
a
(x)
+ a1 (x)
n−1
n
n−1
dx
dx
dx
where an (x), an−1 (x),...,a0 (x) and F (x) depend only on variable x.
If an ODE is not linear, then we call it nonlinear.
Section 1.2 Solutions and Initial Value Problems
A general form for an nth-order equation with variable x and
unknown function y = y (x) can be expressed as
dy
d ny
F x, y , , · · · , n = 0,
dx
dx
(1)
where F is a function that depends on x, y , and the derivatives of
y up to the order n. We assume that the equation holds for all x
in an open interval I (a < x < b, where a or b could be infinite).
d ny
=f
dx n
dy
d n−1 y
x, y , , · · · , n−1
dx
dx
= 0.
(2)
Definition A function ϕ(x) that when substituted by y in
equation (4) or (5) satisfies the equation for all x in the interval I
is called an explicit solution to the equation on I .
Sometimes the solution to the equation is defined implicitly.
Definition A relation G (x, y ) = 0 is said to be an implicit
solution to equation (4) on the interval I if it defines one or more
explicit solutions on I .
Definition A function ϕ(x) that when substituted by y in
equation (4) or (5) satisfies the equation for all x in the interval I
is called an explicit solution to the equation on I .
Sometimes the solution to the equation is defined implicitly.
Definition A relation G (x, y ) = 0 is said to be an implicit
solution to equation (4) on the interval I if it defines one or more
explicit solutions on I .
Example 8 (practice test) Show that the relation y = x + e y is
an implicit solution to the equation (x + y − 1)y ′ = 1.
Definition By an initial value problem for an nth-order
differential equation
d ny
dy
F x, y , , · · · , n = 0
dx
dx
we mean: Find a solution to the differential equation on an interval
I that satisfies at x0 the n initial conditions:
dy
d n−1 y
(x0 ) = y1 , ..., n−1 (x0 ) = yn−1 ,
dx
dx
where x0 ∈ I and y0 , y1 ,...,yn−1 are given constants.
y (x0 ) = y0 ,
Theorem Given the initial value problem
dy
= f (x, y ),
y (0) = y0 ,
dx
assume that f and ∂f /∂y are continuous functions in a rectangle
R = {(x, y ) : a < x < b, c < y < d}
that contains the point (x0 , y0 ). Then the initial value problem has
a unique solution ϕ(x) in some interval x0 − δ < x < x0 + δ, where
δ is a positive number.
Section 1.3 Direction Fields
A first-order equation dy
dx = f (x, y ) specifies a slope at each point
in the xy -plane where f is defined.
A plot of short line segments drawn at various points in the
xy -plane showing the slope of the solution curve there is called a
direction field for the differential equation. Because the direction
field gives the ”flow of solutions”, it facilitates the drawing of any
particular solution (such as the solution to an initial value
problem).
Method of Isoclines
Definition An isocline for the differential equation y ′ = f (x, y ) is
a set of points in the xy -plane where all the solutions have the
same slope dy /dx; thus, it is a level curve for the function f (x, y ).
Section 1.3 Direction Fields
A first-order equation dy
dx = f (x, y ) specifies a slope at each point
in the xy -plane where f is defined.
A plot of short line segments drawn at various points in the
xy -plane showing the slope of the solution curve there is called a
direction field for the differential equation. Because the direction
field gives the ”flow of solutions”, it facilitates the drawing of any
particular solution (such as the solution to an initial value
problem).
Method of Isoclines
Definition An isocline for the differential equation y ′ = f (x, y ) is
a set of points in the xy -plane where all the solutions have the
same slope dy /dx; thus, it is a level curve for the function f (x, y ).
Example 4 (practice test) Draw the isoclines with their direction
markers and sketch the solution to the initial value problem
yy ′ = −x,
y (0) = 4.
Chapter 2. First-Order Differential Equations
Section 2.2 Separable Equations
dy
= g (x)h(y ).
dx
Lets multiply equation (5) by
dy
h(y ) = g (x)dx
R
R dy
g (x)dx
h(y ) =
(3)
dx
h(y )
The solution to an equation (5) is H(y ) = G (x) + C , here H(y )
is an antiderivative of 1/h(y ), G (x) is an antiderivative of g (x), C
is a constant.
Chapter 2. First-Order Differential Equations
Section 2.2 Separable Equations
dy
= g (x)h(y ).
dx
Lets multiply equation (5) by
(3)
dx
h(y )
dy
h(y ) = g (x)dx
R
R dy
g (x)dx
h(y ) =
The solution to an equation (5) is H(y ) = G (x) + C , here H(y )
is an antiderivative of 1/h(y ), G (x) is an antiderivative of g (x), C
is a constant.
Example 6 (practice test) Solve the IVP
√
y dx + (1 + x)dy = 0 y (0) = 1.
Section 2.3 Linear Equations
A linear first-order equation is an equation that can be expressed
in the form
dy
a1 (x)
+ a0 (x)y = b(x),
(4)
dx
where a0 (x), a1 (x), b(x) depend only on x.
We will assume that a0 (x), a1 (x), b(x) are continuous functions of
x on an interval I .
Assume that a1 (x) is never zero on I . We can rewrite (4) in the
standard form
dy
+ P(x)y = Q(x),
(5)
dx
where P(x) = a0 (x)/a1 (x) and Q(x) = b(x)/a1 (x) are continuous
on I .
To solve the first order linear equation
(a) Write the equation in the standard form (5).
(b) Find the integrating factor µ(x) solving differential equation
dµ
− P(x)µ = 0.
dx
(c) Integrate the equation
d
[µy ] = µQ(x)
dx
and solve for y by dividing by µ(x).
To solve the first order linear equation
(a) Write the equation in the standard form (5).
(b) Find the integrating factor µ(x) solving differential equation
dµ
− P(x)µ = 0.
dx
(c) Integrate the equation
d
[µy ] = µQ(x)
dx
and solve for y by dividing by µ(x).
Example 7 (practice test) Find the general solution to the
equation (x 2 − 1)y ′ + 2xy + 3 = 0.
Existence and uniqueness of solution
Existence and uniqueness of solution
Theorem Suppose P(x) and Q(x) are continuous on some
interval I that contains the point x0 . Then for any choice of initial
value y0 , there exists a unique solution y (x) on I to the initial
value problem
y ′ + P(x)y = Q(x),
y (x0 ) = y0 .
(6)
Existence and uniqueness of solution
Theorem Suppose P(x) and Q(x) are continuous on some
interval I that contains the point x0 . Then for any choice of initial
value y0 , there exists a unique solution y (x) on I to the initial
value problem
y ′ + P(x)y = Q(x),
y (x0 ) = y0 .
Example 5 (practice test) Determine (without solving the
problem) an interval in which the solution to the initial value
problem
y ′ + y tan x = sec x, y (0) = 3
is certain to exist.
(6)
Chapter 3. Mathematical methods and numerical methods
involving first order equations.
Section 3.2 Compartmental analysis
The basic one-compartment system consists of a function x(t) that
represents the amount of a substance in the compartment at time
t, an input rate at which the substance enters the compartment,
and an output rate at which the substance leaves the compartment
Because the derivative of x with respect to t can be interpreted as
the rate of change in the amount of the substance in the
compartment with respect to time, the one-compartment system
suggests dx
dt = input rate − output rate as a mathematical model
for the process.
Chapter 3. Mathematical methods and numerical methods
involving first order equations.
Section 3.2 Compartmental analysis
The basic one-compartment system consists of a function x(t) that
represents the amount of a substance in the compartment at time
t, an input rate at which the substance enters the compartment,
and an output rate at which the substance leaves the compartment
Because the derivative of x with respect to t can be interpreted as
the rate of change in the amount of the substance in the
compartment with respect to time, the one-compartment system
suggests dx
dt = input rate − output rate as a mathematical model
for the process.
Mixing problem.
Chapter 3. Mathematical methods and numerical methods
involving first order equations.
Section 3.2 Compartmental analysis
The basic one-compartment system consists of a function x(t) that
represents the amount of a substance in the compartment at time
t, an input rate at which the substance enters the compartment,
and an output rate at which the substance leaves the compartment
Because the derivative of x with respect to t can be interpreted as
the rate of change in the amount of the substance in the
compartment with respect to time, the one-compartment system
suggests dx
dt = input rate − output rate as a mathematical model
for the process.
Mixing problem.
Example 1 (practice test) A large tank initially contains 10 L of
fresh water. A brine containing 20 g/L of salt flows into the tank
at a rate of 3 L/min. The solution inside the tank is kept well
stirred and flows out of the tank at the rate 2 L/min. Determine
the concentration of salt in the tank as a function of time.
Section 3.3 Heating and cooling of buildings.
Let T (t) represent the temperature inside the building at time t
and view the building as a single compartment.
We will consider three main factors that affect the temperature
inside the building.
◮
heat produced by people, lights, and machines inside the
building. This causes a rate of increase in temperature that
we will denote by H(t).
◮
the heating (or cooling) supplied by the furnace (or air
conditioner). This rate of increase (or decrease) in
temperature will be represented by U(t).
◮
the effect of the outside temperature M(t) on the
temperature inside the building.
We have
dT
= K (M(t) − T (t)) + U(t) + H(t)
dt
where H(t) ≥ 0 and U(t) > 0 for furnace heating and U(t) < 0
for air conditioning cooling.
R Kt
T (t) = e−Kt
e [KM(t) + U(t) + H(t)]dt + C
If M(t) = M0 , H(t) = U(t) = 0, and T (0) = T0 , then
T (t) = M0 + (T0 − M0 )e−K (t−t0 ) , here K is a constant.
We have
dT
= K (M(t) − T (t)) + U(t) + H(t)
dt
where H(t) ≥ 0 and U(t) > 0 for furnace heating and U(t) < 0
for air conditioning cooling.
R Kt
T (t) = e−Kt
e [KM(t) + U(t) + H(t)]dt + C
If M(t) = M0 , H(t) = U(t) = 0, and T (0) = T0 , then
T (t) = M0 + (T0 − M0 )e−K (t−t0 ) , here K is a constant.
Example 3. (practice test) An object with temperature 1500 is
placed in a freezer whose temperature is 300 . Assume that the
temperature of the freezer remains essentially constant.
We have
dT
= K (M(t) − T (t)) + U(t) + H(t)
dt
where H(t) ≥ 0 and U(t) > 0 for furnace heating and U(t) < 0
for air conditioning cooling.
R Kt
T (t) = e−Kt
e [KM(t) + U(t) + H(t)]dt + C
If M(t) = M0 , H(t) = U(t) = 0, and T (0) = T0 , then
T (t) = M0 + (T0 − M0 )e−K (t−t0 ) , here K is a constant.
Example 3. (practice test) An object with temperature 1500 is
placed in a freezer whose temperature is 300 . Assume that the
temperature of the freezer remains essentially constant.
1. If the object is cooled to 1200 after 8 min, what will its
temperature be after 18 min?
We have
dT
= K (M(t) − T (t)) + U(t) + H(t)
dt
where H(t) ≥ 0 and U(t) > 0 for furnace heating and U(t) < 0
for air conditioning cooling.
R Kt
T (t) = e−Kt
e [KM(t) + U(t) + H(t)]dt + C
If M(t) = M0 , H(t) = U(t) = 0, and T (0) = T0 , then
T (t) = M0 + (T0 − M0 )e−K (t−t0 ) , here K is a constant.
Example 3. (practice test) An object with temperature 1500 is
placed in a freezer whose temperature is 300 . Assume that the
temperature of the freezer remains essentially constant.
1. If the object is cooled to 1200 after 8 min, what will its
temperature be after 18 min?
2. When will its temperature be 600 ?
Section 3.4 Newtonian mechanics
Mechanics is the study of the motion of objects and the effect of
forces acting on those objects. Newtonian or classical,
mechanics deals with the motion of ordinary objects – that is,
objects that are large compared to an atom and slow moving
compared with the speed of light.
Newton’s laws of motion:
1. When a body is subject to no resultant external force, it
moves with a constant velocity.
2. When a body is subject to one or more external forces, the
time rate of change of the body’s momentum is equal to the
vector sum of the external forces acting on it.
3. When one body interacts with a second body, the force of the
first body on the second is equal in magnitude, but opposite
in direction, to the force of the second body on the first.
Procedure for Newtonian models
1. Determine all relevant forces acting on the object being
studied. It is helpful to draw a simple diagram of the object
that depicts these forces.
2. Choose an appropriate axis or coordinate system in which to
represent the motion of the object and the forces acting on it.
3. Apply Newton’s second law to determine the equations of
motion for the object.
g = 32 ft/sec2 = 9.81 m/sec2
An object of mass m is given an initial downward velocity v0 and
allowed to fall under the influence of gravity. Assuming the
gravitational force is constant and the force due to air resistance is
proportional to the velocity of the object with some coefficient b
(b¿0), determine the equation of motion for this body.
A model for the velocity of the falling body is expressed by the IVP
v (0) = v0
m dv
dt = mg − bv ,
mg −bt/m
mg
e
+ v0 −
v (t) =
b
b
b
mg m
−m
t
x(t) = mg
)
b t + b v0 − b (1 − e
An object of mass m is given an initial downward velocity v0 and
allowed to fall under the influence of gravity. Assuming the
gravitational force is constant and the force due to air resistance is
proportional to the velocity of the object with some coefficient b
(b¿0), determine the equation of motion for this body.
A model for the velocity of the falling body is expressed by the IVP
v (0) = v0
m dv
dt = mg − bv ,
mg −bt/m
mg
e
+ v0 −
v (t) =
b
b
b
mg m
−m
t
x(t) = mg
)
b t + b v0 − b (1 − e
Example An object of mass 5 kg is given the initial downward
velocity of 50 m/sec and then allowed to fall under the influence of
gravity. Assume that the force in newtons due to air resistance is
−10v , where v is the velocity of the object in m/sec.
An object of mass m is given an initial downward velocity v0 and
allowed to fall under the influence of gravity. Assuming the
gravitational force is constant and the force due to air resistance is
proportional to the velocity of the object with some coefficient b
(b¿0), determine the equation of motion for this body.
A model for the velocity of the falling body is expressed by the IVP
v (0) = v0
m dv
dt = mg − bv ,
mg −bt/m
mg
e
+ v0 −
v (t) =
b
b
b
mg m
−m
t
x(t) = mg
)
b t + b v0 − b (1 − e
Example An object of mass 5 kg is given the initial downward
velocity of 50 m/sec and then allowed to fall under the influence of
gravity. Assume that the force in newtons due to air resistance is
−10v , where v is the velocity of the object in m/sec.
1. Determine the equation of motion of the object.
An object of mass m is given an initial downward velocity v0 and
allowed to fall under the influence of gravity. Assuming the
gravitational force is constant and the force due to air resistance is
proportional to the velocity of the object with some coefficient b
(b¿0), determine the equation of motion for this body.
A model for the velocity of the falling body is expressed by the IVP
v (0) = v0
m dv
dt = mg − bv ,
mg −bt/m
mg
e
+ v0 −
v (t) =
b
b
b
mg m
−m
t
x(t) = mg
)
b t + b v0 − b (1 − e
Example An object of mass 5 kg is given the initial downward
velocity of 50 m/sec and then allowed to fall under the influence of
gravity. Assume that the force in newtons due to air resistance is
−10v , where v is the velocity of the object in m/sec.
1. Determine the equation of motion of the object.
2. After how many seconds will the velocity of the object be 70
m/sec?