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Section 1.1
Differential Equations
& Mathematical Models
Differential Equations – “Equations with derivatives in them.”
Examples:
1.
dy
 xy  e
x
dx
2.
3x y 
4x
2
1 y
1
3. y″ – 2 = 3x ln(x) + y2
What is a solution to a differential equation?
A differential equation will often have infinitely many solutions.
For example, here are some of the many solutions of
dy
dx
1. y = x2
2. y = x2 + e–x
3. y = x2 + 4e–x
4. y = x2 + 31429674e–x
:
:
“Family of solutions”
 y  x  2x
2
Calculus Review:
If
dy
dx
 f (x)
then y 

f ( x ) dx
A differential equation will usually have infinitely many solutions, but there
times when a differential equation will have only one solution or no solutions.
Example:
(y ′)2 + y2 = –1
Some differential equations will have solutions, but unfortunately we can't
write them down (in terms of our elementary functions).
Example:
y  e
x
2
General Solutions vs. Particular Solutions
1. Solve
dy
 3x
2
 3x
2
 3x
2
dx
2. Solve
3. Solve
dy
dx
dy
dx
if y = 5 when x = 1.
, y(1) = 5.
Initial Value Problem –
consists of a differential equation along with an initial condition y(xo) = yo
Definition: Order
The order of a differential equation is the order of the highest derivative
appearing in it.
Expressing differential equations:
Often we will be able to express 1st order differential equations as
dy
dx
 f ( x , y ).
Expressing differential equations:
We will always be able to express. . . .
1st order differential equations in the form F(x, y, y′) = 0
2nd order differential equations in the form F(x, y, y′, y″) = 0
:
nth order differential equations in the form F(x, y, y′, y″, y″′, . . . . , y(n)) = 0
Definition: Solution to a Differential Equation
A function u(x) is a solution to the differential equation F(x, y, y′, y″, . . , y(n)) = 0
on an interval J if u, u′, u″, . . . , u(n) exist on J and F(x, u, u′, u″, . . . , u(n)) = 0
for all x on J.
Ex. 1
1
(a) Show that y(x) = 1/x is a solution to 3 xy   2 y 
on the interval [1, 20].
x
Ex. 1
1
(b) Show that y(x) = 1/x is not a solution to 3 xy   2 y 
on the interval [-20, 20].
x
Ex. 2
(a) Show that y1(x) = sin(x) is a solution to (y ′ )2 + y2 = 1
(b) Show that y2(x) = cos(x) is a solution to (y ′ )2 + y2 = 1
Partial Derivatives
Ordinary Differential Equations vs. Partial Differential Equations
Section 1.2
Integrals as General & Particular Solutions
Ex. 1 Solve
dy
dx
 co s x
Ex. 2 Solve x
dy
dx
1  2x 1
2
Ex. 3 Solve x
dy
dx
 1  2 x  1,
2
y (1)  3
Position - Velocity – Acceleration
s(t) = position
s′ (t) = velocity
Force = Mass x Acceleration
s″ (t) = acceleration
Ex. 5 A lunar lander is falling freely toward the surface of the moon at a speed of
450 m/s. Its retrorockets, when fired, provide a constant deceleration of 2.5 m/s2
(the gravitational acceleration produced by the moon is assumed to be included in
the given deceleration). At what height above the lunar surface should the retro
rockets be activated to ensure a "soft touchdown" (velocity = 0 at impact)?
Ex. 5 A lunar lander is falling freely toward the surface of the moon at a speed of
450 m/s. Its retrorockets, when fired, provide a constant deceleration of 2.5 m/s2
(the gravitational acceleration produced by the moon is assumed to be included in
the given deceleration). At what height above the lunar surface should the retro
rockets be activated to ensure a "soft touchdown" (velocity = 0 at impact)?
Section 1.3
Slope Fields & Solution Curves
Slope field for
dy
dx
 co s x
Slope field for
dy
dx
 2x
Ex. 1 Sketch the slope field for y′ = –x
Ex. 2 Sketch the slope field for y′ = x2 + y2
2
2
Ex. 3 Examine some solution curves of y  x
dy
dx
 0.
On the following slope field, draw the solution curve which satisfies the
initial condition of. . . . .
(a) y(2) = –1
(b) y(–1) = 3
(c) y(0) = 0
(d) y(0) = 1
Calculus Review (definition of continuity):
f (x) is continuous at xo if lim f ( x )  f ( x o )
x  xo
Calculus Review (definition of continuity):
f (x) is continuous at xo if lim f ( x )  f ( x o )
x  xo
f (x, y) is continuous at (xo, yo) if
lim
 x , y    xo , y o 
f  x , y   f  xo , yo 
Theorem I: Existence & Uniqueness of Solutions
Suppose that f (x, y) is continuous on some rectangle in the xy-plane containing
the point (xo, yo) in its interior and that the partial derivative fy is continuous on
that rectangle. Then the initial value problem dydx  f ( x , y ); y ( x o )  y o , has a
unique solution on some open interval Jo containing the point xo.
Ex. 4 Determine what this theorem says about the solutions in each of the
following differential equations:
(a)
dy
dx
 x  3y ,
2
5
y (2)  4
Ex. 4 Determine what this theorem says about the solutions in each of the
following differential equations:
(b) y 2  x 2 d y  0,
dx
y (0 )  1
Ex. 4 Determine what this theorem says about the solutions in each of the
following differential equations:
(c)
y  x
2
2
dy
dx
 0,
y (1)  0
Ex. 4 Determine what this theorem says about the solutions in each of the
following differential equations:
(d) dy 
dx
1
x y
,
y (2)  2
Ex. 4 Determine what this theorem says about the solutions in each of the
following differential equations:
(e) dy 
dx
1
x y
,
y (3)  5
Ex. 4 Determine what this theorem says about the solutions in each of the
following differential equations:
(f)
dy
dx

7
x 
7
y,
y (3)  0
Section 1.4
Separable Equations & Applications
Definition: Separable Differential Equation
A first order differential equation
dy
dx
 f ( x, y )
is said to be separable if
f (x, y) can be written as a product of a function of x and a function of y
(i.e.
dy
dx
 g ( x)h( y )
).
Definition: Separable Differential Equation
A first order differential equation
dy
dx
 f ( x, y )
is said to be separable if
f (x, y) can be written as a product of a function of x and a function of y
(i.e.
dy
 g ( x)h( y )
).
dx
Examples:
1.
2.
3.
dy
dx
  3 x  1  7 y  2 y  3 
2
y

e
1
2
  sin x  x  

dx
y

5


dy
dy
dx
x  3x  6
2

y  y  10
5
dy
g (x)

To solve a separable differentiable equation of the form
dx
h( y)
we proceed as follows:
dy
g (x)

To solve a separable differentiable equation of the form
dx
h( y)
we proceed as follows:
dy
dx

g (x)
h( y)
h(y) dy = g(x) dx
dy
g (x)

To solve a separable differentiable equation of the form
dx
h( y)
we proceed as follows:
dy
dx

g (x)
h( y)
h(y) dy = g(x) dx
 h ( y ) dy   g ( x ) dx
(Then integrate both sides and solve for y, if this is possible.)
Ex. 1 Solve
dy
dx
 4x y
3
2
Ex. 2 Solve
dy
dx
 x y
Justification for why this method for solving separable differentiable
equations actually works.
Ex. 3 Solve
dy
dx
  6 xy ,
y (0 )   7
Ex. 4 Solve
dy
dx

x
y
Review of general solutions and particular solutions.
Definition: Singular Solution
A particular solution to a first order differential equation is said to be a
singular solution if it does not come from the general solution.
Ex. 5 Solve
dy
dx
 6 x  y  1
2 /3
Ex. 5 Solve
dy
dx
 6 x  y  1
2 /3
In general:
If we have the differential equation
dy
dx
 g ( x)h( y )
, and h(y) has a zero of yo
then the function y(x) = yo will be a singular solution.
Applications
If y changes at a rate proportional to y then
Radioactive material & half-lives
dy
dx
 ky
(for some constant k).
Ex. 6 A radioactive substance has a half-life of 5 years. Initially there are
128 grams of this substance. How much remains after t years?
Ex. 7 The half-life of radioactive cobalt is 5.27 years. Suppose that a
nuclear accident has left the level of cobalt radiation in a certain region at
100 times the level acceptable for human habitation. How long will it be
until the region is again habitable?
Carbon dating
Ex. 8 Carbon extracted from an ancient skull contained only one-sixth as
much 14C as carbon extracted from present-day bone. How old is the skull?
Newton's law of cooling (heating)
According to Newton's law of cooling, the time rate of change of the temperature
T of a body immersed in a medium of constant temperature A is proportional to
the difference T – A . That is:
dT
dt
 k T  A .
Ex. 9 A cake is removed from an oven at 210° F and left to cool at a room
temperature, which is 70° F. After 30 min the temperature of the cake is 140° F.
When will it be 100° F?
Section 1.5
Linear First-Order Equations
Definition: Linear Differential Equation (First Order)
A first order differential equation is linear if there are functions P(x) and Q(x) so that
dy
dx
Examples:
1. y′ + sin(x) y = ex
2. y′ – sin(x) y = ex
3. y′ = 3x2y + x3 – 4x + 1
4. cos(x) y′ – sec(x) y = x2
 P (x) y  Q (x)
Definition: Linear Differential Equation (First Order)
A first order differential equation is linear if there are functions P(x) and Q(x) so that
dy
dx
Integrating Factor:
P ( x ) dx

e
 P (x) y  Q (x)
Steps you MUST show when solving a 1st order linear differential equation:
1. Put the differential equation in the form:
dy
 P (x) y  Q (x)
dx
2. Compute μ.
3. Multiply μ on both sides of the differential equation to obtain

dy
dx
4. Write this as (μy)′ = μ Q(x)
5. Solve this last differential equation via integration.
 P ( x) y  Q ( x)
Ex. 1 Solve:
dy
dx
 x y
Why isn't there a “+C” in the integrating factor μ?
Ex. 2 Solve:
x y′ + 2y = 10x3;
y(1) = 5
Reminder: (Theorem I from section 1.3)
Suppose that f (x, y) is continuous on some rectangle in the xy-plane
containing the point (xo, yo) in its interior and that the partial derivative fy is
continuous on that rectangle. Then the initial value problem
dy
dx
 f ( x , y );
y ( xo )  yo ,
has a unique solution on some open interval Jo containing the point xo.
Theorem I
If the functions P(x) and Q(x) are continuous on the open interval J
containing the point xo, then the initial value problem
dy
 P ( x ) y  Q ( x );
dx
has a unique solution y(x) on J .
y ( xo )  yo
Ex. 3 Solve: y  
y
x
 2;
y (2)  4
Reminder of a result from calculus:
dx
dy

1
dy
dx
Applications
Mixture Problems
Solutes, Solvents, & Solutions
Q = amount of solute, V = volume of solution
ri = rate in ro = rate out, ci = concentration in, co = concentration out
Ex. 4 Consider a large tank holding 1000 L of water into which a brine
solution of salt begins to flow at a constant rate of 6 L/min. The solution inside
the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min.
If the concentration of salt in the brine entering the tank is 1 kg/L, determine
when the concentration of salt in the tank will reach 0.5 kg/L.
Ex. 5 For the mixture problem described in example 4, assume now that the brine
leaves the tank at a rate of 5 L/min instead of 6 L/min and assume that the tank starts out
with a concentration of 0.1 kg/L (everything else stays the same as it was in example 4
though). Determine the concentration of salt in the tank as a function of time.
Ex. 6 A swimming pool whose volume is 10,000 gallons contains water that
is 0.01% chlorine. Starting at t = 0, city water containing 0.001% chlorine is
pumped into the pool at a rate of 5 gal/min, and the pool water flows out at the
same rate. What is the percentage of chlorine in the pool after 1 hr? When
will the pool be 0.002% chlorine?
Section 1.6
Substitution Methods & Exact Equations
Ex. 1 Solve
dy
dx
 9 x  y  4
2
If a differential equation can be written as
dy
 f ( A x  B y  C ),
dx
let u = Ax + By + C and the resulting differential equation in terms of u and x
will be separable.
Definition: First Order Homogenous Differential Equation
dy
 y
 f  
A 1st order differential equation that can be expressed as
dx
 x
is said to be homogenous.
Examples:
1.
2
 y
 y
    3   7
dx  x 
 x
dy
2.
 y  5y
y   sin    2
x
 x
3.
 y  5x
y   sin    2
y
 x
4.
x
2
2
dy
dx

x  y
2
2
If a differential equation can be written as
 y
 f  ,
dx
 x
dy
let
u 
y
x
and the resulting differential equation in terms of u and x will be separable.
Ex. 2 Solve x 2
dy
dx
 xy  x e
2
 y/x
Here's why a 1st order homogenous differential equation can be turned into a
y
separable differential equation by the substitution u  .
x
Here's why a 1st order homogenous differential equation can be turned into a
y
separable differential equation by the substitution u  .
x
 y
 f  
dx
 x
dy
Here's why a 1st order homogenous differential equation can be turned into a
y
separable differential equation by the substitution u  .
x
 y
 f  
dx
 x
dy
d
dx
 ux  
f (u )
Here's why a 1st order homogenous differential equation can be turned into a
y
separable differential equation by the substitution u  .
x
 y
 f  
dx
 x
dy
d
dx
x
 ux  
du
dx
f (u )
 u  f (u )
Here's why a 1st order homogenous differential equation can be turned into a
y
separable differential equation by the substitution u  .
x
 y
 f  
dx
 x
dy
d
dx
x
 ux  
du
f (u )
 u  f (u )
dx
du
dx

f (u )  u
x
Definition: Bernoulli Differential Equation
dy
n
 P (x) y  Q (x) y
order differential equation that can be expressed as
dx
is said to be a Bernoulli differential equation.
A 1st
Examples:
1.
dy
dx
 3 xy  sin  x  y
1  3x
8
2.
y  4e y 
3.
x y   5 tan  x  y  sinh  x  y
x
2
5  2x
y
10
5
Ex. 3 Solve
dy
dx

2
x
y  (8 x  5) y
3
4
Ex. 3 Solve
dy
dx

2
x
y  (8 x  5) y
3
4
Definition: Exact Differential Equation
A 1st order differential equation that can be expressed as M ( x , y )  N ( x , y )
dy
dx
with
is said to be an exact differential equation.
0
Exact Differential Equation
M ( x, y )  N ( x, y )
dy
0
dx
M(x, y) dx + N(x, y) dy = 0
Exact Differential Equation
Solving
M(x, y) dx + N(x, y) dy = 0
Ex. 4 Solve (4x – y) dx + (6y – x) dy = 0
  2 xy  3 x
2
Ex. 5 Solve
dy
dx

2x y  4 y
2
2
3

Section 1.8
Acceleration-Velocity Models
Ex. 1 Suppose that an object is a distance of ho from the surface of the earth
when it is given an initial velocity of vo. Determine s(t), the position of the
object expressed as a function of time. (Assume that the only force acting on
the object is due to gravity. Assume a constant acceleration due to gravity of g).
Ex. 2 Suppose that an object is a distance of ho from the surface of the earth
when it is given an initial velocity of vo. Determine s(t), the position of the
object expressed as a function of time, this time including air resistance.
(Use g for the acceleration due to gravity and assume that the force caused by
the air resistance is proportional to the velocity.)
Ex. 3 An object of mass 3 kg is released from rest 500 m above the ground
and allowed to fall under the influence of gravity. Assume the force due to
air resistance is proportional to the velocity of the object with a drag
coefficient of 4/3 sec–1. Determine when the object will strike the ground.
Newton's Law of Gravitation –
The gravitational force of attraction between two point masses M and m
located at a distance r apart is given by F 
GMm
r
2
.
Ex. 4 A lunar lander is falling freely toward the surface of the moon at a
speed of 450 m/s. Its retrorockets, when fired, provide a constant deceleration
of 2.5 m/s2 . At what height above the lunar surface should the retro rockets be
activated to ensure a soft touchdown (velocity = 0 at impact)?

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