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MTH 253
Calculus (Other Topics)
Chapter 9 – Mathematical Modeling with
Differential Equations
Section 9.1 – First-Order Equations
and Applications
Copyright © 2006 by Ron Wallace, all rights reserved.
Terminology …
Differential Equation
 An equation involving one or more
derivatives of an unknown function.
Examples:
2
d y
dy
3  2y  0
2
dx
dx
y ''' 2 y '  cos x
Terminology …
Order of a Differential Equation
 The highest order derivative in the
equation.
Examples:
2
d y
dy
3  2y  0
2
dx
dx
Order = 2
y ''' 2 y '  cos x
Order = 3
Terminology …
Solution of a Differential Equation
 A function whose derivatives make the
differential equation a true statement.
 A general solution is a function w/
parameters that represents ALL possible
solutions.
Example:
dy
x
ye
dx
A solution:
General solution:
y  xe
x
y  xe  Ce
x
NOTE: # of constants in the general solution = order of the DifEq
x
Terminology …
 Initial-Value Problem
 A differential equation with conditions
that determine a unique solution.
Example:
dy
x
ye
dx
y (0)  3
General solution:
y  xe  Ce
x
3  0e  Ce
0
0
C 3
IVP solution:
y  xe  3e
x
NOTE: # of initial conditions of an IVP = order of the DifEq
x
x
A Differential Equations Course
The development and study of
methods for solving differential
equations and IVP’s.
 Categorizing Differential Equations
 Explicit methods
 Find solutions and general solutions
 Numerical methods
 Find a set of ordered pairs that
approximate the solution of an IVP.
 Applications
1st Order Separable Equations
 The variables can algebraically be
separated, giving the form …
h( y )dy  g ( x)dx
Example:
y dy
2
 y cos x
x dx

1
dy  x cos x dx
y
NOTE: Since we multiplied by x and divided by y2,
we must assume that x0 and y0.
Solving 1st Order Separable Equations
1. Algebraically separate the variables.
2. Integrate both sides of the equation.
3. Solve for y (if possible).
Example:

y dy
 y 2 cos x
x dx
1
 y dy   x cos x dx
 ye
x sin x  cos x C



1
dy  x cos x dx
y
ln y  x sin x  cos x  C
y  Ke x sin x cos x
Check the solution?
1st Order Linear Equations
 Equations that can be algebraically
manipulated into the form …
dy
 p( x) y  q ( x)
dx
Example:
dy
x
ye
dx

p ( x)  1
q( x)  e
x
Solving 1st Order
Linear Equations
dy
 p( x) y  q ( x)
dx
p ( x ) dx

1. Find ............   e
(with c = 0)
2. Find ............     q ( x)dx

3. Solution is ... y  
Example:
 dx

 e
 e x
dy
x
ye 
dx
   e x e x dx   dx  x  c
y  e x ( x  c)
 is called an integrating factor.
Applications: Mixing Problems
 A tank contains a solution with a known amount of
a substance (y).
 A solution with a known concentration of the
substance is entering the tank at a known rate.
 The tank is draining at a known rate.
 The solution in the tank is kept thoroughly stirred.
 How much of the substance is in the tank at any
point in time?
dy
 (substance in) - (substance out)
dt
 (concentration in)(rate in) -
y
(rate out)
solution amt in tank at time t
Initial Condition: y(0)  y0 (initial amt of substance in tank)
Applications: Mixing Problems
A polluted lake contains 1 lb of mercury salts
per 100,000 gallons of water. The volume of the
lake is 560,000 gallons. Water is pumped from
the lake at 1000 gallons/hour and replaced by
fresh water a the same rate. How much mercury
salts is in the lake after 100 hours of pumping?
dy
y
y
 0
(1000)  
dt
560, 000
560

y  ce
t
560
 5.6e
t
560

1
1
 y dy    560 dt
y (100)  4.68 lbs.
Initial Condition: y (0)  5.6 lbs.
Applications: Free Falling Object
 An object is in free fall.
 At time t = 0, the height is s0 and velocity is v0.
 Two forces are acting on the object:
 Gravity: FG = -mg (mass times gravity [32 ft/sec2])
 Air Resistance: FR = -cv (v is the object’s velocity)
 FR is called drag
 c is a positive constant depending on the shape of
the object
 Newton’s Second Law of Motion (F = ma)
d 2s
m 2  FG  FR  mg  cv
dt

Initial Condition: v(0)  v0
dv c
 v  g
dt m