Transcript Calculus I - Parkland College

Section 5.1
First-Order Systems & Applications
Suppose x and y are both functions of t. Solve:
x′ = 3x – y
y′ = 2x + y – et
Why would we consider such things?
Ex. 1 Give the system of diff eqs which describe the following tanks
containing brine solution:
Theorem:
Consider the following system of diff eqs (all xi, pij, and fi are functions of t )
x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
x2′ = p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2
x3′ = p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3
:
:
xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn
Let J be an open interval containing t = a. Suppose the functions pij and the
functions fk are continuous on J. Then the system of differential equations has a
unique solution that satisfies the following initial conditions:
x1(a) = b1 , x2(a) = b2 , x3(a) = b3 , ......... , xn(a) = bn
If we have a system of higher order diff eqs, we can transform it into a
system of first order diff eqs.
Ex. 3 Rewrite the one diff eq x(3) + 3x″ + 2x′ – 5x = sin(2t) as a system of
first order diff eqs.
Ex. 4 Rewrite the following system as a system of first order diff eqs.
2x″ = –6x + 2y
y″ = 2x – 2y + 40sin(3t)
Section 5.2
The Method Of Elimination
Review of solving systems of algebraic equations (2 equations with 2 unknowns)
i. Substitution method
ii. Elimination method
iii. Cramer's rule
Ex. 1 Find the general solution to the following system by using a variant of
the substitution method.
x′ = 4x – 3y
y′ = 6x – 7y
Ex. 1 Find the general solution to the following system by using a variant of
the substitution method.
x′ = 4x – 3y
y′ = 6x – 7y
Ex. 2 Find the general solution to the following system by using a variant of
the elimination method.
x′ = 4x – 3y
y′ = 6x – 7y
Ex. 2 Find the general solution to the following system by using a variant of
the elimination method.
x′ = 4x – 3y
y′ = 6x – 7y
We shall now use the following notation:
L will be a linear operator of the form
L = anDn + an–1Dn–1 + an–2Dn–2 + ⋯ + a2D2 + a1D + a0
Ex. 3 Find the general solution to the following general system of diff eqs:
L1x + L2y = f1(t)
L3x + L4y = f2(t)
Ex. 3 Find the general solution to the following general system of diff eqs:
L1x + L2y = f1(t)
L3x + L4y = f2(t)
Ex. 4 Find the general solution to x′ = 2x + y
y′ = 2x + 3y + e5t
(D–2)x – y = 0
–2x + (D–3)y = e5t
Solution:
D2
1
2
D 3
x
1
D2
1
D 3
2
D3
0
e
5t
[(D–2)(D–3) – 2] x = (D – 3)0 + e5t
y
D2
0
2
e5t
[(D–2)(D–3) – 2] y = (D – 2) e5t + 2(0)
(D2–5D+4) x = e5t
(D2–5D+4) y = 5e5t – 2e5t
(D–4)(D–1) x = e5t
(D–4)(D–1) y = 3e5t
xc = c1e4t + c2et
xp = Ae5t
⇒
yc = c3e4t + c4et
xp = (1∕4)e5t
x = c1e4t + c2et + (1∕4)e5t
yp = Be5t
⇒
yp = (3∕4)e5t
y = c3e4t + c4et + (3∕4)e5t
Ex. 4 Find the general solution to x′ = 2x + y
y′ = 2x + 3y + e5t
Solution:
x = c1e4t + c2et + (1∕4)e5t
y = c3e4t + c4et + (3∕4)e5t
Plugging these solutions into the first equation in the initial problem we get:
x′ = 2x + y
(c1e4t + c2et + (1∕4)e5t)′ = 2(c1e4t + c2et + (1∕4)e5t) + (c3e4t + c4et + (3∕4)e5t)
4c1e4t + c2et + (5/4)e5t = 2c1e4t + 2c2et + (1/2)e5t + c3e4t + c4et + (3/4)e5t
0 = (–2c1 + c3)e4t + (c2 + c4)et
–2c1 + c3 = 0
c2 + c4 = 0
⇒
c3 = 2c1
c4 = –c2
Section 5.3
Matrices & Linear Systems
x1′ = 2x1 + x2
x2′ = 2x1 + 3x2
x1′ = 2x1 + x2
x2′ = 2x1 + 3x2
⇒
 2 1
x  
x

 2 3
Previously we would state:
The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et
x2′ = 2x1 + 3x2
x2 = 2c1e4t – c2et
Previously we would state:
The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et
x2′ = 2x1 + 3x2
x2 = 2c1e4t – c2et
We now would state:
 2 1
The general solution to the system x  
x turns out to be:

 2 3
 e 4t 
 et 
x  c1  4t   c2  t 
 2e 
 e 
Previously we would state:
The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et
x2′ = 2x1 + 3x2
x2 = 2c1e4t – c2et
We now would state:
 2 1
The general solution to the system x  
x turns out to be:

 2 3
 e 4t 
 et 
x  c1  4t   c2  t 
 2e 
 e 
We shall find that general solutions to these "2x2 homogenous systems" will take
this form of x  c1 x (1)  c2 x (2) (where x (1) and x (2) are two linearly independent
vectors).
x1′
x2′
x3′
:
xn′
= p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
= p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2
= p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3
:
= pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn
x1′
x2′
x3′
:
xn′
= p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
= p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2
= p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3
:
= pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn

x  Px  f
x1′
x2′
x3′
:
xn′
= p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
= p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2
= p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3
:
= pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn
This system of diff eqs is said to be homogenous if f  0

x  Px  f
x1′
x2′
x3′
:
xn′
= p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
= p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2
= p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3
:
= pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn

x  Px  f
This system of diff eqs is said to be homogenous if f  0
Thus, the matrix equation would be x  Px for a homogenous system.
We shall now see many theorems very similar to theorems and definitions we
had in chapter 2.
Definition:
x (1) , x (2) , x (3) , .... , x ( n)are said to be linearly independent if the equation
c1 x (1)  c2 x  c3 x (3) 
 cn x ( n)  0 only has the solution of c1 = c2 = ⋯ = cn = 0.
Definition:
If x (1) , x (2) , x (3) , .... , x ( n) are solutions to x  Px then we define the
Wronskian of these solution to be
W  x (1) , x (2) , x (3) , .... , x ( n ) 
 x (1) , x (2) , x (3) , .... , x ( n )
x1(1)
x1(2)
x1(3)
x1( n )
x2(1)
x2(2)
x2(3)
x2( n )
 x3(1)
x3(2)
x3(3)
x3( n )
xn(1)
xn(2)
xn(3)
xn( n )
Theorem:
Suppose x (1) , x (2) , x (3) , .... , x ( n) are solutions to x  Px on an interval J
where all the pij functions are continuous.
(a) If x (1) , x (2) , x (3) , .... , x ( n) are linearly dependent then W = 0 for every
point on the interval J.
(b) If x (1) , x (2) , x (3) , .... , x ( n) are linearly independent then W ≠ 0 for every
point on the interval J.
Theorem:
Suppose x (1) , x (2) , x (3) , .... , x ( n) are linearly independent solutions to x  Px
on an interval J where all the pij functions are continuous. The general
solution to x  Px is given as:
x  c1 x (1)  c2 x  c3 x (3) 
 cn x ( n)
Theorem:
Suppose x p is a particular solution to the nonhomogenous system x  Px  f
and xc is the general solution to the corresponding homogenous system x  Px .
Then the general solution to the nonhomogenous system x  Px  f is x  x p  xc .
Ex. 1
x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
(a) Write this system as a matrix equation.
Ex. 1
x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
(b) Verify that the following three
vectors are solution vectors:
x (1)
et 
 
  0 ,
et 
 
x (2)
 3et 
 
  2et  ,
 et 
 
x (3)
 e 2t 


 3e2t 
 e 2t 


Ex. 1
x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
(c) Verify that these are
linearly independent vectors.
x (1)
et 
 
  0 ,
et 
 
x (2)
 3et 
 
  2et  ,
 et 
 
x (3)
 e 2t 


 3e2t 
 e 2t 


Ex. 1
x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
(d) Give the general solution for the system.
Ex. 1
x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
(e) Give the general solution for x1.
Give the general solution for x2.
Give the general solution for x3.
Ex. 1
x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
(f) Solve the initial value problem:
x′1 = x1 + x2 – 2x3
x1(0) = 9,
x′2 = –x1 + 2x2 + x3
x′3 =
x2 – x3
x2(0) = –2, x3(0) = 5
Section 5.4
The Eigenvalue Method for
Homogeneous Systems
Review of eigenvalues and eigenvectors:
Let A be a square matrix. The vector x is said to be an eigenvector for the
eigenvalue λ if Ax   x .
Review of eigenvalues and eigenvectors:
Let A be a square matrix. The vector x is said to be an eigenvector for the
eigenvalue λ if Ax   x .
What is the connection between eigenvectors and solutions to systems of diff
eqs?
Ex. 1 We previously (in section 5.3) found three linearly independent solution
 1 1 2
 e 2t 
vectors to the system x    1 2 1  x.
One of these was 3e 2t  .




2t
 0 1 1
e 


 1 1 2 
Verify that this solution is an eigenvector of  1 2 1  .


 0 1 1
 Ae kt 


Ex. 2 Suppose that  Bekt  (where A, B, C, and k are constants) is a solution
 Ce kt 


vector to the system x  Px . Show that this solution vector is an eigenvector
of the matrix P.
Theorem:
Given a homogenous system x  Px , suppose λ is an eigenvalue of P with
t
eigenvector v . Then ve is a nontrivial solution vector to x  Px .
Ex. 3 Use eigenvalues/eigenvectors to find the general solution to the
following system and write your final solution in scalar form.
x′1 = 3x1 + x2
x′2 = 3x1 + 5x2
Ex. 3 Use eigenvalues/eigenvectors to find the general solution to the
following system and write your final solution in scalar form.
x′1 = 3x1 + x2
x′2 = 3x1 + 5x2
Note:
Suppose x  Px where P is an nxn matrix. If there are n distinct real
eigenvalues of P then we've got n linearly independent solution vectors. This
means we can write down the general solution as
x  c1v1e1t  c2v2e2t  c3v3e3t   cnvnent
(here the λi are the eigenvalues, the vi are the eigenvectors and the ci are
arbitrary constants).
If we have less than n distinct eigenvalues, or if some of the eigenvalues are
complex then we will run into trouble and need to do something else.
Theorem:
Given a homogenous system x  Px , suppose α + βi is a complex
eigenvalue of P with eigenvector a  bi  a  bi . Then the real and
 i
a  bi will form two linearly
imaginary parts of the vector e
independent solution vectors.


Theorem:
Given a homogenous system x  Px , suppose α + βi is a complex
eigenvalue of P with eigenvector a  bi  a  bi . Then the real and
 i
a  bi will form two linearly
imaginary parts of the vector e
independent solution vectors.


Note that there are formulas for these two vectors, but you should not use
them!
v1  et a cos   t   b sin   t 
v2  et


 a sin   t   b cos   t 
Instead of using these formulas we shall just use this theorem which indicates
that we should find the one vector e   i a  bi , then split it up into its real
and imaginary parts.


Ex. 4 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –x1 – x2
x′2 = 4x1 – x2
Ex. 4 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –x1 – x2
x′2 = 4x1 – x2
Section 5.5
Multiple Eigenvalue Solutions
If we have less than n distinct eigenvalues (where P is an nxn matrix) then we
may have a hard time finding enough linearly independent solution vectors.
When an eigenvalue, say λ, is a repeated root with multiplicity k of the
characteristic equation, then two possibilities arise:
If we have less than n distinct eigenvalues (where P is an nxn matrix) then we
may have a hard time finding enough linearly independent solution vectors.
When an eigenvalue, say λ, is a repeated root with multiplicity k of the
characteristic equation, then two possibilities arise:
1. There are "just enough" linearly independent eigenvectors associated with
 : v1 , v2 , v3 , , vk . Then there are "just enough" solution vectors:
v1et , v2et , v3et , , vk et for us to form the general solution to the diff eqs.
2. There are not enough linearly independent eigenvectors associated with λ.
In this case λ is said to be defective.
We shall cover the first of these two cases.
Ex. 1 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –13x1 + 40x2 – 48x3
x′2 = –8x1 + 23x2 – 24x3
x′3 =
3x3
Ex. 1 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –13x1 + 40x2 – 48x3
x′2 = –8x1 + 23x2 – 24x3
x′3 =
3x3