Math 220, Differential Equations
• Professor Charles S.C. Lin
• Office: 528 SEO, Phone: 413-3741
• Office Hours: MWF 2:00 p.m. & by
appointments
• E-mail address: cslin@uic.edu
1
Teaching Assistant
•
•
•
•
•
Mr. Diego Dominici
Office: 607 SEO
Office Hours: ?
Phone: 996-4814
e-mail: ddomin1@uic.edu
2
Please check:
• www.math.uic.edu/~berger/M220/index.ht
ml for Syllabus, assignments, etc...
• www.awlonline.com/nagle for interactive
CD
3
Differential Equation:
Classifications
• Ordinary differential equations, order!
• Partial differential equations
• Linear equations: i.e. linear in the dependent
variable(s).
• Nonlinear differential equations: not linear
• For example: xy" y ' xy  0
This eq. appears in stress analysis,
aerodynami cs.
4
Explicit solution and Implicit solution
• If a function satisfies a differential equation, for
example:
3
y   x  8 , satisfy th e D.E.
dy 3 x 2

which can be verified easily.
dx 2 y
• Such a function, defined explicitly as a function
of independent variable x is called an explicit
solution. On the other hand, the equation
x  y  e xy  0, satisfies the D.E.
5
Given by
• the following:
(*)
• The equation
dy
(1  ye xy )

.
xy
dx
(1  xe )
x ye  0
xy
• is said to defined an implicit solution of the
equation (*) above.
6
In fact, there are many solutions to a D.E
such as (*) above.
• To find a solution passing through a specific point
in xy-plane, we need to impose a condition,
known as : initial value, i.e. y(x0) = y0. This is
known as the initial value problem. We shall
assume that the function f(x,y) is sufficiently
smooth, that a solution always exists. Namely:
• Theorem (Existence and uniqueness): The I.V.P.
y'  f ( x, y) , y( x0 )  y0
• always has a unique solution in a rectangle
containing the point (x0, y0), if f and f x are
continuous there.
7
Direction Fields
• Consider the first order D. E.
dy
 f ( x, y)
dx
• the equation specifies a slope at each point
in the xy-plane where f is defined.
• It gives the direction that a solution to the
equation must have at each point.
8
A plot of short line segments
drawn at various points in the xyplane showing the slope of the
solution curve this is called a “
direction field ” for the
differential equation.
• The direction field gives us the “ flow of
solutions ”.
9
Example
• For the equation
dy
 x 2  y.
dx
•
•
•
•
Using Maple
with(DEtools);
eq:=diff(y(t),t)=t^2-y(t);
DEplot(eq,y(t),t=-5..5,y=-5..5,arrows=slim)
10
Using Maple program, we have
the following
• graph:
11
Another example:
• Consider the logistic equation for the
population of a certain species:
dp
 3 p  2 p 2 , with p(0)  2 .
dt
•
•
•
•
using maple, we write in commands:
eq:=diff(p(t),t)=3*p(t)-2*p(t)^2;
DEplot(eq,p(t), t=0..5,p=0..5,arrows=slim);
and get
12
Its direction field
• like this:
13
The Method of Isoclines
• Consider the differential equation
• (*)
dy/dx = f(x,y).
• The set of points in the xy-plane where all
the solutions have the same slope dy/dx;
i.e. the level curves for the function f(x,y)
are called the isoclines for the D. E. (*).
• This is the family of curves f(x,y) = C.
• This gives us a way to draw direction field.
14
Example
• For the differential equation
•
•
•
•
y'  x  y
f(x,y) = x + y, and the set of points where:
x + y = c, are straight lines with slope (-1).
We can now draw the isoclines for the D. E.
and the solution passing through a given
initial point can also be drawn.
15
Let us graph the isoclines of
f(x,y) = x + y.
• and compare it to the direction field of it, we
see
16
Maple example
• Let us consider the IVP for y’ = x^2 - y,
with three sets of initial points: [0,-1], [0,0]
and [0,2]. What will be the corresponding
solutions?
17
Separable Equation
• Given a differential equation
dy
 f ( x, y)
dx
• If the function f(x,y) can be written as a
product of two functions g(x) and h(y), i.e.
• f(x,y) = g(x) h(y), then the differential eq. is
called separable.
18
Example
• The equation
dy 3x  xy

2
dx
1 y
• is separable, since
 3 y 
3x  xy
2  ( x )
2   g ( x ) h( y )
1 y
 1 y 
19
Method for solving separable
equation
• Separable equation can be solved easily,
• Rewrite the equation:
dy
 g ( x ) h ( y ) in the form
dx
dy
 g ( x ) dx then find
h( y )
antiderivatives on both sides. i. e.
dy
 h( y )   g ( x ) dx  C.
20
Example
• Consider the initial value problem
dy
y 1

dx
x3
y ( 1)  0
21
Using Maple:
• we can solve the IVP with the following
Maple commands.
• ODE:=diff(y(x),x)=(y(x)-1)/(x-3);
• IC:=y(-1)=0;
• IVP:={ODE,IC};
• GSOLN:=dsolve(ODE,y(x));
• Then use the IC to find the arbitrary
constant.
22
Linear Equations
• We shall study how one can solve a first order
linear differential equation of the form:
dy
a1 ( x )  a0 ( x ) y  b( x ),
dx
• We first rewrite the above equation in the so
called “standard form”:
y ' P( x ) y  Q( x ).
23
Integration Factor
• Suppose we multiply a function (x) to the
above equation, we get:
( x) y' ( x) P( x) y  ( x)Q( x)
• Is it possible for us to find (x) such that the
left hand side
• ?
d
LHS 
[  ( x) y]
dx
24
• Since
d
dy
[  ( x ) y ]   ( x )   '( x ) y
dx
dx
• We see that this can be done, if
 ' ( x )   ( x ) P( x ), which implies
d
 P( x )dx , by integration, we get

ln|  ( x )|   P( x )dx. Hence
 ( x)  e
 P ( x ) dx
.
25
In this case,
• we can solve it by integration.
• Note that:
[  ( x ) y ]'   ( x )Q( x ), implies
 ( x ) y    ( x )Q( x )dx  c, consequently
we have
1
y( x) 
 ( x)
  ( x)Q( x)dx  c.
26
Examples:
• Consider the D.E.
y ' y  e .
3x
• Solution:
• Another example: solve the following initial
value problem:
dx
3
2
t
 3t x  t , x (2)  0.
dt
27
Application: Mixing Problems
(Compartmental Analysis)
• 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
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• Let x(t) be the mass of salt in the tank at
time t. The rate at which salt enters the tank
is equal to “input rate - output rate”. Thus
dx
x(t )
 (6 l / min)( 1kg / l )  (6 l / min)(
kg / l )
dt
1000
or we have :
dx
3x
 6
, with x(0)  0
dt
500
29
The equation is separable
• We can solve it easily, using the initial
condition, we get
x (t )  1000(1  e

3t
500
). Thus, the concentration
of salt in the tank at time t is:
3t
x (t )
1

500
 1 e
kg / L. When will this =
?
1000
2
3t
1

500
Set 1  e
 , we find t = 115.52 min.
2
30
Existence and Uniqueness
Theorem
• Suppose P(x) and Q(x) are continuous on
the interval (a,b) that contains the point x0.
Then the initial value problem:
• y + P(x)y = Q(x), y(x0)=y0
• for any given y0.
• has a unique solution on (a,b).
31
Application to Population
Growth
• If we assume that the growth rate of a population is
proportional to the population present, then it leads
to a D.E.:
• Let p(t) be the population at time t. Let k > 0 be
the proportionality constant for the growth rate and
let p0 be the population at time t = 0. Then a
• mathematical model for a population could be:
dp
 kp,
dt
p( 0)  p0
32
This can be solved easily.
• Example: In 1790 the population of the
United States was 3.93 million, and in 1890
it was 62.95 million. Estimate the U.S.
population as a function of time.
33
Application to Newtonian Mechanics
• The study of motion of objects and the effect of
forces acting on those objects is called Mechanics.
A model for Newtonian mechanics is based on
Newton’s laws of motion: Let us consider an
example: An object of mass m is given an initial
velocity of 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 .
Determine the equation of motion for this object.
34
Solution
• Since the total force acting on the object is
• F = FG - FA = mg - k v(t). And according to
Newton’s 2nd law of motion, F = m a, we see
that
•
m a = mg - k v.
• Let x(t) be the position function of the object at
time t, and
•
v(t) = dx/dt, a = dv/dt.
35
Equation of motion can be
rewritten as:
• The following separable initial value problem.
dv
k
 g
v,
dt
m
v (0)  v0 , where
• We can solve the equation easily, and obtain:
mg
mg
v(t ) 
 (v0 
)e
k
k
kt

m
.
36
Now, to find the position function
x(t)
• Suppose that at t = 0, the object is x0 units
above the ground, i.e. x(0) = x0 . Then for
the position function x(t), we have the
following I.V.P.
dx mg
mg

 (v0 
)e
dt
k
k
kt

m
, with x(0)  x0 .
• This can be solved easily.
37
We obtain:
• The equation of motion:
kt

m
mgt m
mg
x(t ) 
 (v0  )(1  e )  x0 .
k
k
k
38
Linear Differential Operators (4.2)
• We shall now consider linear 2nd order
equations of the form:
2
d y
dy
(*) a2 ( x) 2  a1 ( x)
 a0 ( x) y  b( x),
dx
dx
Where ai ( x), and b( x) are continuous in x on
some interval I. We generally assume that a2 ( x)
is never zero on I, thus (*) can be put in the form
d2y
dy
()
 p ( x)
 q ( x) y  g ( x),
2
dx
dx
The so called " Standard form".
39
Homogeneous equation
associated with ()
• is the equation
2
d y
dy
 p ( x)  q ( x) y  0,
2
dx
dx
in the sence of " linear algebra". For simplicity
we shall define L[y]  LHS of this equation, i.e.
L[y] : y" p(x)y' q(x)y .
L is called a differenti al operator of order 2.
40
Remark on linearity of the operator L,
and linear combinations of solutions to
homogeneous equation.
• We have L[y1+y2]= L[ y1]+ L[y2],
• for any constants  and , and any twice
differentiable functions y1 and y2 .
• Theroem1. If y1 and y2 are solutions of the
homogeneous equation
• (HE): y+py+qy=0, then any linear
combination y1+y2 of y1 and y2 is also a
solution of (HE).
41
Consider an example
• L[y] = y + 4y + 3y, We use the
convention Dy = y , D2y = y , Dny = y(n) ,
• and rewrite L[y] = D2y + 4Dy +3y, or
symbolically, L = D2 + 4D +3. Since
formerly D2 + 4D +3 = (D + 3)(D + 1), we
see that:
• L[y] = (D + 3)(D + 1)[y]. The solutions for
• L[y] = 0 are y1 = e-x , and y2 = e -3x.
42
Existence and Uniqueness of 2nd
order equation
• Theorem 2. Let p(x), q(x) and g(x) be
continuous on an interval (a,b), and x0 (a,b).
Then the I.V.P.
y" p( x) y ' q( x) y  g ( x) ;
y ( x0 )  y0 , y ' ( x0 )  y1 ,
has a unique solution on the whole interval
(a, b) for any choice of the initial values
y0 and y1 .
43
Fundamental Solutions of
Homogeneous Equations
• Let us first define the notion of the Wronskian of
two differentiable functions y1 and y2. The
function
W [ y1 , y2 ]( x) : y1 ( x) y2 ' ( x)  y1 ' ( x) y2 ( x)
is called the Wronskian of y1 and y2 .
(in honor of H. Wronski, 1778 - 1853).
Remark : this can be viewed as a 2x2
determinan t.
44
Fundamental solution set
• A pair of solutions [y1, y2] of L[y] = 0, on (a,b)
where L[y] = y+py+qy is called a fundamental
solution set, if W[y1, y2](x0)  0 for some x0
(a,b) . A simple example: Consider
• L[y] = y+9y. It is easily checked that y1 = cos 3x
and y2 = sin 3x are solutions of L[y] = 0. Since the
corresponding Wronskian W[y1, y2](x) = 3  0 ,
thus {cos 3x, sin 3x} forms a fundamental solution
set to the homogenenous eq: y  + 9y = 0. We see
that any linear combination c1 y1 + c2 y2 also
satisfies L[y] = 0. This is known as a general 45
solution
Linear Independence, Fundamental
set and Wronskian
• Theorem. Let y1 and y2 be solutions to the equation
y + py + qy = 0 on (a,b). Then the following
statements are equivalent:
• (A) {y1, y2} is a fundamental solution set on (a,b).
• (B) y1 and y2 are linearly independent on (a,b).
• (C) The W[y1, y2](x) is never zero on (a,b).
• For the proof, we need some linear algebra, i.e.
• Linearly dependent vectors,uniqueness theorem
46
etc...
Reminder
• First Hour Exam:
• Date: June 15 (Friday)
• Room: TBA
47
Homogeneous Linear Equations
With Constant Coefficients
•
•
•
•
•
Recall: For equations of the form
ay + by + cy = 0,
by subsituting y = e r x, we obtain the
auxiliary eq: ar2 + br + c = 0. If r1 and r2
are two distinct roots, then a general solution is
of the form y = c1exp(r1x)+ c2exp(r2x), where
c1 and c2 are arbitrary constants.
48
Repeated Roots
•
•
•
•
•
•
•
If in the above equation, r1 = r2 = r, then a
general solution is of the form
y = c1exp(rx) + c2x exp(rx),
Example: consider the D.E. : y + 4y´ + 4 = 0.
Its auxiliary equation is: r2 + 4r + 4 = 0, hence
r = -2 is a double root, the general solution is
y = c1e -2x + c2x e -2x,
49
Cauchy-Euler Equations
• If an equation is of the form:
ax2y + bxy´ + cy = h(x), a, b, c are constants,
then by letting x = e t, we transform the original
equation into:(with t as the independent
variable), ay  + (b-a)y´ + cy = h(e t). An
equation with constant coefficients. Hence can
be solved by the method of constant
coefficients. The equation above is known as a
Cauchy-Euler Equation.
50
Reduction of order
• We know ,in general, a second order linear
differential equation has two linearly independent
solutions. If we already have one solution, how can
we find the other one?
• Let f(x) be a solution to y  + p(x)y´ + q(x)y = 0.
• We will try to find another solution of the form
• y(x) = v(x)f(x), with v(x) a non-constant function.
• Formerly, we have y ´ = v´f + vf ´, and y  = …
• set w = v ´, etc…, we obtain a separable eq. in w.
51
• Finally find v from w by integration.
Example
•
•
•
•
•
•
Given f(x) = x-1 is a solution to
x2 y  - 2xy´ -4y = 0, x > 0;
find a second linearly independent solution.
First write the D.E. in standard form.
Next compute v.
Finally, 2nd independent solution is y = v f.
52
Auxiliary Eq. With Complex Roots
• If the auxiliary equation of a linear 2nd order D.
E. with constant coefficents: ar2 + br + c = 0,
has complex roots, (when b2 - 4ac < 0 ). i.e.
• r1 =  + i and r2 =  - i, where  and  are
real numbers, then the solutions are
• y1 = e ( + i)x , and y2 = e ( - i)x. Since we know
that e i x = cos  x + i sin  x , we simply take
• y1 = e x cos  x , and y2 = e x sin  x as the
two linearly independent solutions.
53
And the general solution is of the
form
• y(x) = c1 e x cos  x + c2 e x sin  x, where
c1 and c2 are arbitrary constants.
• Remark about complex solution:
• z(x) = u(x) + iv(x) to L[z] = 0 and the fact
that in this case, we also have L[u] = 0 and
L[v] = 0. Thus the real part and the
imaginary part of a complex solution to
L[y] = 0 are also solutions of L[y] = 0.
54
Example
• Consider the D.E.
z"6 z '10 z  0. Clearly th e auxiliary
equation is : r 2  6r  10  0. Since
b  4ac  (6)  4(1)(10)  4  0, we see
2
2
6 4
that the roots are r 
 3  i.
2
3x
3x
Hence y1  e cos x, and y 2  e sin x. and
the general solution is y  c1e cos x  c2 e sin x.
3x
3x
55
Nonhomogeneous Equation And the
the method of Superposition
• Let L be a linear operator of 2nd order, i.e.
L= D2 + pD + q, and g  0. The equation:
• L[y] = g, is called a Nonhomogeneous eq.
• We wish to solve the equation L[y] = g ,
using a particular solution to L[y] = g , and
a fundamental solution set to L[y] = 0. First
let me introduce the concept of the method
of superposition.
56
Theorem: Let y1 be a solution to the
equation L[y] = g1, and let y2 be a
solution to the equation L[y] = g2,
where g1 and g2 are two functions.
Then for any two constants c1 and c2,
the linear combination c1 y1 + c2 y2
is a solution to the equation
L[y] = c1 g1 + c2 g2 . (This is known as
the Superposition principle).
57
Proof
58
Representation Theorem of L[y] = g.
• Theorem: Let yp(x) be a particular solution to the
nonhomogeneous equation (*) L[y] = g(x),
where L[y]= y  + p(x) y  + q(x) y , on the
interval (a,b) and let y1(x) and y2(x) be a
fundamental solution set of L[y] = 0 on the
interval (a,b). Then every solution of (*) can be
written in the form
• (**) y(x) = yp(x) + c1 y1(x) +c2 y2(x) . This is
known as the general solution to (*).
59
Example
• Given that yp(x) = x2 is a particular solution
of the equation:
• (*)
y  - y = 2 - x2,
• find a general solution of (*).
• Note the auxiliary equation is r 2 - 1 = 0. It
follows that a general solution of (*) is of
the form y = x2 + c1e-x + c2e x.
60
Superposition Principle & the
Method of Undetermined
coefficients.
• Example: Find a general solution to the D.E.
y"3y  x  e .
2
x
• Step 1: We first consider the associated
homogenous equation:
y"3 y  0
61
Step 2: Find particular solution to
the Non-homogenous equation
using the Superposition Principle
• There are “2” equations:
y"3 y  x
2
y"3 y   e .
x
62
To find a particular solution to
each of the above equations
• We use the method of undetermined
coefficients, that is: for the first equation, we try
yp = ax2 + bx + c, and
• For the second equation, we try yp = Aex.
• If any term in the trial expression for yp is a
solution to the corresponding homogeneous
equation, then we replace yp by x yp, etc…. See
table 4.1 on Page 208 of your book.
63
Next we present a more general
method, known as:
• The method of variation of parameters,
64
The Method of Variation of
Parameters
• Consider the non-homogenous linear second
order differential equation :
y" p( x ) y ' q ( x ) y  g ( x ). ..........(1)
Let { y1 ( x ), y2 ( x )} be a fundamental solution
set for the corresponding homogeneous equation
y" p( x ) y ' q ( x ) y  0. We try particular solution
of the equation (1) in the form:
y p  v1 ( x ) y1 ( x )  v2 ( x ) y2 ( x ).
65
Where v1 and v2 are functions to
be determined. We obtain:
• two equations (by avoiding 2nd order derivatives
for the unknows and from L[yp]=g):
v1 ' y1  v2 ' y2  0, and
v1 ' y1 ' v2 ' y2 '  g. solving these two
equations simultaneuously, we get
 g ( x ) y2 ( x )
 g ( x ) y1 ( x )
v1 '( x ) 
, v2 '( x ) 
.
W[ y1 , y2 ]( x )
W[ y1 , y2 ]( x )
66
Where v1 and v2 are functions to
be determined. We obtain:
• two equations (by avoiding 2nd order derivatives
for the unknows and from L[yp]=g):
v1 ' y1  v2 ' y2  0, and
v1 ' y1 ' v2 ' y2 '  g. solving these two
equations simultaneuously, we get
 g ( x ) y2 ( x )
 g ( x ) y1 ( x )
v1 '( x ) 
, v2 '( x ) 
.
W[ y1 , y2 ]( x )
W[ y1 , y2 ]( x )
67
Finally, solution is found by
integration.
• Example:
Find a general solution t o the D.E.
y"4 y '4 y  e
2 x
ln x .
68
69