ODE
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Section 1.1: Some Basic Mathematical Models; Direction Fields
A direction field (or a slope field) is a graph consisting of short line segments that indicate the
slopes of the tangent lines to the family of curves of a differential equation at a discrete number
of points.
An equilibrium solution is a solution to a differential equation that does not change (i.e., is a
constant). Equilibrium solutions are found by setting y = 0 (because the derivative of a constant
is zero) and solving the resulting equation.
A stable equilibrium solution is a solution that other solutions tend to converge toward.
Direction field lines point toward stable equilibriums.
An unstable equilibrium solution is a solution that other solutions tend to diverge away from.
Direction field lines point away from unstable equilibriums.
More comments on direction fields:




dy
 f t, y  .
dt
f is sometimes called the rate function (think of the exponential growth differential
dy
 ky , where k is the growth rate (or rate costant)).
equation
dt
Direction fields are nice because they do not require solving the differential equation.
Use a computer to do the work!
Direction fields are useful in studying equations of the form
Section 1.2: Solutions of Some Differential Equations
dy
b
 ay  b have a general solution y    ce at
Equations are of the form
dt
a
Notes:
 As we solve these equations by performing a single integration, a single constant of
integration will always be generated.
 To get a concrete value for the constant of integration, a desired point that the solution
should contain must be given.
 The given point on the curve is called an initial condition.
 A differential equation combined with an initial condition is called an initial value
problem (IVP).The solution obtained with the arbitrary constant of integration (due to no
specified initial condition) is called the general solution.
 All the graphs of the solutions for all possible values of the constant of integration are
called the integral curves.
 The solution to the equations in this section will be found by manipulating the equation
so that one side is an exact differential and then integrating.
 This will be analogous to the technique of “separation of variables” you may have
learned from section 7.2 of Smith & Minton’s 3rd edition of Calculus: Early
Transcendental Functions.
ODE
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Section 1.3: Classification of Differential Equations
Ordinary vs. Partial Differential Equations
Ordinary Differential Equations
 Only ordinary derivatives appear in the
equation
Example:
d 2Q  t 
dQ  t  1
L
R
 Q t   E t 
2
dt
dt
C
(RLC series circuit with a drive voltage)
Partial Differential Equations
 Only partial derivatives appear in the
equation
Example:
 2u  x, t  u  x, t 
2

x 2
t
(heat conduction equation)
Systems of Differential Equations
 …consist of multiple differential equations involving multiple variables.
 Used when one quantity depends on other quantities…
dx
 ax   xy
dt
 Example: Lotka-Volterra (predator-prey) equations:
dy
 cy   xy
dt
Order of a Differential Equation
 …is the order of the highest derivative that appears (analogy: the degree of a polynomial
is the largest exponent)
 General form for an nth order differential equation: F t , u  t  , u  t  , , u  n  t   0 for



some function F.
We’ll deal with differential equations that can be solved explicitly for the highest
derivative: y  n  t   f t , y  t  , y  t  , , y  n1  t  .


Linear vs. Nonlinear Differential Equations
 F t , y  t  , y  t  , , y  n  t   0 is linear if F is a linear function of y  t  , y  t  ,



, y  n t  ;
note: F does not have to be linear in t…
The general form of a linear differential equation of order n is:
an  t  y  n  t   an1  t  y  n1  t    a1  t  y  t   a0 t  y t   g t 


Nonlinear differential equations are not of the above form…
Many times, nonlinear equations can approximated as being linear over small regions of
the solution. This process is known as linearization.
Solutions of Differential Equations
 …are functions  such that all necessary derivatives exist over an interval of interest of
n
n 1
the independent variable and that satisfy     t   f t ,   t  ,    t  , ,     t  .



To verify a given solution, substitute it and its derivatives into the given differential
equation to show that the equation is satisfied.
ODE
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Section 2.1: Linear Equations; Method of Integrating Factors
dy
 p t  y  g t 
Form of a linear first-order differential equation:
dt
Formula for the solution of a first-order linear differential equations with constant
coefficients:
t
dy
 ay  g  t  , then   t   eat , and y  t   e at  eas g  s  ds  ce at
If
dt
t0
Formula for the solution of a general first-order linear differential equation:
t

dy
1 
 p  t  y  g  t  , then   t   exp   p   dt  , and y  t  
If

s
g
s
ds

c








dt
  t   t0

Section 2.2: Separable Equations
dy
 f  x, y 
dx
M  x
Separable first-order differential equation: f  x, y   
N  y
General form of a first-order differential equation:
Another form for a separable first-order differential equation: M  x  dx  N  y  dy  0
Solution:
M  x
dy

dx
N  y
 N  y  dy   M  x  dx
dy
y
 f  v  , where v  can be converted to a separable equation.
dx
x
Recall the following homework problem:
Note: equations of the form
ODE
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Section 2.3: Modeling with First Order Equations
Three Key Steps in Mathematical Modeling
 Construct the model (use the steps at the end of section 1.1).
 Analyze the model by solving it or making appropriate simplifications to solve it.
 Compare the results from the model with experiment and observation.
Section 2.4: Differences Between Linear and Nonlinear Equations
Linear Fist-Order Differential Equations
dy
 p t  y  g t 
dt
Solution:
First, compute   t   exp   p  t  dt 
Then answer is explicitly
1 
y t  
  t  g  t  dt  c 
 t  
Theorem 2.4.1: Existence and Uniqueness of
the Solution of a Linear First-Order
Differential Equation
If the functions p and g are continuous in an
open interval I containing the point t = t0, then
there exists a unique function y = (t) in I that
satisfies the initial value problem
y  p  t  y  g  t  and y  t0   y0 .
Nonlinear Fist-Order Differential Equations
dy
 f t, y 
dt
… we have only looked at one kind, which is
separable equations:
dy
 F  y  G t 
dt
Solution:
dy
First, separate variables:
 G  t  dt
F  y
Then integrate for an implicit equation
dy
 F  y    G t  dt
Theorem 2.4.2: Existence and Uniqueness of
the Solution of a Nonlinear First-Order
Differential Equation
f
If the functions f and
are continuous in the
y
rectangle containing the point (t0, y0), then
there exists a unique solution y = (t) in that
rectangle that satisfies the initial value problem
y  f  t , y  and y  t0   y0 .
The proof is rooted in the fact that we can write
the solution in terms of operations on
The proof is beyond the scope of this class. It
continuous functions.
should be noted that if the conditions of the
hypothesis are not satisfied, then it is possible
to get multiple solutions for one initial
condition
Note that if the uniqueness criteria are met, then graphs of two solutions can not intersect,
because if they did then that implies two solutions (i.e., contradicting uniqueness) for a given
initial condition.
Interval of Definition:
Interval of Definition:
…is determined by any discontinuities in p and …cannot be determined by looking at f.
g
… must look at the solution itself
General Solutions:
General Solutions:
…can always be written using the constant of
…many times cannot be summarized with the
ODE
Test #1 Review
integration c.
Implicit vs. Explicit Solutions:
…the solution is stated explicitly as an
integral.
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constant of integration; example: equilibrium
solutions.
Implicit vs. Explicit Solutions:
…the solutions obtained are usually implicit.
Section 2.5: Autonomous Equations and Population Dynamics
In this section, we look at a restricted form of a nonlinear separable equation:
dy
 f  y .
dt
When f is independent of t, the equation is called autonomous (i.e., t does not drive the solution;
the solution only depends on itself…).
Logistic Growth Model
 Accounts for the fact that the growth rate depends on population  instead of
dy
 ry , use

dt
dy
 h  y  y for some function h.

dt
 We like h to model the fact that as populations get larger, the growth rate gets
smaller due to “using up” the environement.  h  y   r  ay is a simple model
that gets used extensively.
dy
  r  ay  y …called the logistic equation, and is usually written the form:

dt
r
dy
y

 r 1   y where K  is the environmental carrying capacity (or

a
dt
 K
saturation level), and r is the intrinsic growth rate.
dy
 0 , we are also solving
 When finding equilibrium solutions by solving
dt
f  y   0 ; zeros of f are called critical points.




A phase line graph shows the y-axis and how solutions either converge toward or
diverge from equilibrium solutions.
dy
For stable equilibrium solutions,
points toward the equilibrium.
dt
dy
For unstable equilibrium solutions,
points away from the equilibrium.
dt
dy
y

  r 1   y , T is called the threshold level for the
For he equation
dt
 T
population.
ODE
Test #1 Review
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Section 2.6: Exact Equations and Integrating Factors
In this section, we examine the equation M  x, y   N  x, y 
If once again the solution comes from   x, y   c , then
dy
0.
dx
d
d
  x, y    c
dx
dx
 dx  dy

0
x dx y dx
dy
 x  x, y    y  x, y   0
dx
Thus:
Theorem 2.6.1: Exact Differential Equations and Their Solution
Let the functions M , N , M y , and N x be continuous in some rectangular region R. Then the
equation M  x, y   N  x, y  y  0 is an exact differential equation in the region R if and only if
M y  x, y   N x  x, y  at each point of R. That is, there exists a function  satisfying
 x  x, y   M  x, y  and  y  x, y   N  x, y  if and only if M y  x, y   N x  x, y  .
The method of solving an exact equation is identical to finding the potential for a conservative
vector field…
Integrating Factors…
Sometimes a differential equation that is not exact can be converted to an exact equation by
multiplying the equation by an integrating factor   x, y  :
dy
0
dx
dy
  x, y  M  x, y     x , y  N  x , y   0
dx
For this equation to be exact, it must be true that
  M  y    N x
M  x  N  y
Which leads to the partial differential equation:
 y M    x N   N x
 y M  x N   M y  N x    0
If  can be found, then an exact equation is formed, whose solution can be obtained using the
prior technique. Unfortunately, the equation to find  is usually harder to solve than the original
equation (in general).
However, we can look for simplifications, like what happens if the original equation is
such that  depends only on x or y.
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Test #1 Review
Differential Equation where the integrating
factor  depends on x only (i.e.,     x  ):
  M  y    N x
d
N   Nx
dx
d  M y  Nx


dx
N
M y 
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Differential Equation where the integrating
factor  depends on y only (i.e.,     y  ):
  M  y    N x
d
M  M y   Nx
dy
d  Nx  M y


dy
M
Criterion for     x  only:
Criterion for     y  only:
M y  Nx
Nx  M y
is a function of x only.
N
Note that the equation for  is both linear and
separable now.
is a function of y only.
M
Note that the equation for  is both linear and
separable now.
Section 2.7: Numerical Approximations: Euler’s Method
y   f  x, y 
y 
y
x
y
 f  x, y 
x
y  f  x, y  x
yi 1  yi  f  xi , yi  x
yi 1  yi  f  xi , yi  x
Or, letting h = x, yi 1  yi  hyi
To perform Euler’s method quickly on your calculator:
 Store the initial conditions in variables X and Y.
 Enter the update formula for Y then for X separated by a colon.
 Repeatedly Use 2ND ENTER until you are at the final x-value.
 If you want to see the intermediate values of Y, you’ll have to type Y every time…
 Example: for y  x  e y , y(1) = -2, h = 0.25
1X
-2  Y
Y + (X + e^(-Y))*0.25  Y:X+0.25X
Y
2ND ENTER 2ND ENTER …