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Mathematics

MATH 251

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Systems of First Order Linear Differential Equations
We will now turn our attention to solving systems of simultaneous
homogeneous first order linear differential equations. The solutions of such
systems require much linear algebra (Math 220). But since it is not a
prerequisite for this course, we have to limit ourselves to the simplest
instances: those systems of two equations and two unknowns only. But first,
we shall have a brief overview and learn some notations and terminology.
A system of n linear first order differential equations in n unknowns (an n ×
n system of linear equations) has the general form:
x1′ = a11 1 a x +12 2 a x + g 1n n 1
x2′ = a21 1 a x +22 2 a x + g 2n n 2
x3′ = a31 1 a x +32 2 a x + g 3n n 3 (*)
: : :
: : :
x ′ = a x + a x + … + a x + g
n n1 1 n2 2 nn n n
Where the coefficients a ijs, andg i’s are arbitrary functions of t. If every
term g iis constant zero, then the system is said to be homogeneous.
Otherwise, it is a nonhomogeneous system if even one of the g’s is nonzero.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 1 The system (*) is most often given in a shorthand format as a matrix▯vector
equation, in the form:
x′ = Ax + g
′
x1 a 11 a 12 ... a1n x1 g1
x ′ a a ... a x g
2 21 22 2n 2 2
x ′ a 31 a 32 ... a3n x3 g3
3 = +
: : : : : : :
: : : : : :
:
x n′ a 1 a n2 ... a nn xn gn
x′ A x g
Where the matrix of coefficients, A, is called the coefficient matrix of the
system. The vectors x′, x, and g are
′
x1 1 g1
x ′ x2 g2
2
x 3′ x3 g3
x′ = , x = : , g = : .
:
x ′ n gn
n
For a homogeneous system, g is the zero vector. Hence it has the form
x′ = Ax.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 2 Fact: Every n▯th order linear equation is equivalent to a systemof n first
order linear equations.
Examples:
(i) The mechanical vibration equation m + γu′ + ku = F(t) is equivalent to
x′= x
1 2
x′= − k x − γ x + F(t)
2 m 1 m 2 m
Note that the system would be homogeneous (respectively,
nonhomogeneous) if the original equation is homogeneous
(respectively, nonhomogeneous).
(ii) y″′ − 2y″ + 3y′ − 4y = 0 is equivalent to
x1′ = x2
x2′ = x3
x3′ = 4x1− 3x +22x 3
© 2008, 2012 Zachary S Tseng D▯1 ▯ 3 This process can be easily generalized. Given an n▯th order linear equation
a y + a y (n−1+ a y(n−2+ … + a y″ + a y′ + a y = g(t).
n n−1 n−2 2 1 0
Make the substitutions: x = y, x = y′, x = y″, … , x = y (n−, and x ′ = y .
1 2 3 n n
The first n − 1 equations follow thusly. Lastly, substitute the x’s into the
original equation to rewrite it into the n▯th equation and obtain the system of
the form:
x ′ = x
1 2
x 2 = x 3
x 3 = x 4
: : :
: : :
x n−1′ = x n
x ′ = −a 0 x − a1 x − a2 x −...− a n−1x + g(t)
n a 1 a 2 a 3 a n a
n n n n n
Note: The reverse is also true. Given an n × n system of linear equations, it
can be rewritten into a single n▯th order linear equation. (The result is not
unique. There are multiple ways to do this.)
© 2008, 2012 Zachary S Tseng D▯1 ▯ 4 Exercises D▯1.1:
1 – 3 Convert each linear equation into a system of first order equations.
1. y″ − 4y′ + 5y = 0
2. y″′ − 5y″ + 9y = t cos 2t
3. y(4)+ 3y″′ − πy″ + 2πy′ − 6y = 11
4. Rewrite the system you found in (a) Exercise 1, and (b) Exercise 2, into a
matrix▯vector equation.
5. Convert the third order linear equation below into a system of 3 first
order equation using (a) the usual substitutions, and (b) substitutions in the
reverse order: x = y″, x = y′, x = y. Deduce the fact that there are multiple
1 2 3
ways to rewrite each n▯th order linear equation into a linear system of n
equations.
y″′ + 6y″ + y′ − 2y = 0
Answers D▯1.1:
1. x ′ = x 2. x ′ = x
1 2 1 2
x2′ = −5x 1 4x 2 x2′ = x3
x3′ = −9x1+ 5x 3 tcos 2t
3. x1′ = x2
x2′ = x3
x3′ = x4
x4′ = 6x 1 2πx +2πx − 33 + 114
0 1 0 0
0 1
4. (a) x′ = x (b) x′ = 0 0 1 x + 0
−5 4
−9 0 5 tcos2t
5. (a) x1′ = x2 (b) x1′ = −6x1− x 2 2x 3
x2′ = x3 x2′ = x1
x ′ = 2x − x − 6x x ′ = x
3 1 2 3 3 2
© 2008, 2012 Zachary S Tseng D▯1 ▯ 5 A Crash Course in (2 × × 2) Matrices
Several weeks worth of matrix algebra in an hour… (Relax, we will only
study the simplest case, that of 2 × 2 matrices.)
Review topics:
1. What is a matrix (pl. matrices)?
A matrix is any rectangular array of numbers (called entries). Each
entry’s position is addressed by the row and column (in that order)
where it is located. For example, a52 represents the entry positioned at
the 5th row and the 2nd column of the matrix A.
2. The size of a matrix
The size of a matrix is specified by 2 numbers
[number of rows] × [number of columns].
Therefore, an m × n matrix is a matrix that contains m rows and n
columns. A matrix that has equal number of rows and columns is
called a square matrix. A square matrix of size n × n is usually
referred to simply as a square matrix of size (or order) n.
Notice that if the number of rows or columns is 1, the result (respectively, a
1 × n, or an m × 1 matrix) is just a vector. A 1 × n matrix is called a row
vector, and an m × 1 matrix is called a column vector. Therefore, vectors are
really just special types of matrices. Hence, you will probably notice the
similarities between many of the matrix operations defined below and vector
operations that you might be familiar with.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 6 3. Two special types of matrices
Identity matrices (square matrices only)
The n × n identity matrix is often denoten by I
1 0 0
1 0 0 1 0
I2= 0 1 , 3 = , etc.
0 0 1
Properties (assume A and I are of the same size):
AI = IA = A
Inx = x, x = any n × 1 vector
Zero matrices – matrices that contain all▯zero entries.
Properties:
A + 0 = 0 + A = A
A0 = 0 = 0A
4. Arithmetic operations of matrices
(i) Addition / subtraction
a b e f a ± e b ± f
± =
c d g h c ± g d ± h
© 2008, 2012 Zachary S Tseng D▯1 ▯ 7 (ii) Scalar Multiplication
a b ka kb
k c d = kc kd , for any constant k.
(iii) Matrix multiplication
a b e f ae +g af + bh
=
c d g h ce +g cf + dh
The matrix multiplication AB = C is defined only if there are as many
rows in B as there are columns in A. For example, when A is m × k
and B is k × n. The product matrix C is going to be of size m × n, and
whose ij▯th entcy, is equal to the vector dot product between the i▯
th row of A and the j▯th column of B. Since vectors are matrices, we
can also multiply together a matrix and a vector, assuming the above
restriction on their sizes is met. The product of a 2 × 2 matrix and a 2▯
entry column vector is
a b ax by
= .
c d cx dy
Note 1: Two square matrices of the same size can always be
multiplied together. Because, obviously, having the same number of
rows and columns, they satisfy the size requirement outlined above.
Note 2: In general, AB ≠ BA. Indeed, depending on the sizes of A
and B, one product might not even be defined while the other product
is.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 8 5. Determinant (square matrices only)
For a 2 × 2 matrix, its determinant is given by the formula
a b
det = ad −c
c d
Note: The determinant is a function whose domain is the set of all
square matrices of a certain size, and whose range is the set of all real
(or complex) numbers.
6. Inverse matrix (of a square matrix)
Given an n × n square matrix A, if there exists a matrix B (necessarily
of the same size) such that
AB = BA = I,
n
−1
then the matrix B is called the inverse matrix of A, denoted A . The
inverse matrix, if it exists, is unique for each A. A matrix is called
invertible if it has an inverse matrix.
a b
Theorem: For any 2 × 2 matrix A = ,
c d
its inverse, if exists, is given by
d −b
−1 1
A = ad −bc −c a .
Theorem: A square matrix is invertible if and only if its
determinant is nonzero.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 9 −2 2 −3
Examples: Let A 2 and B =1 4 .
1 −2 − 2 − 3= 2 − − − −3)
(i) 2A − B2=5 2 −1 4 10 − −) 4−4
0 −1
=
11 0
1 −2 2 −3 2 + − −8 4 −11
(ii) AB = = =
5 2 1 4 0 − −15 + 8 − 7
On the other hand:
2 −3 − 2 2 −5 − − 6 − 13 −10
BA = = =
−1 4 2 + 20 2 + 19 10
(iii) det(A) = 2 − (−10)det(B) = 8 − 3 = 5.
Since neither is zero, as a result, they are both invertible matrices.
1 2 2 1 2 2 1/6 1/6
(iv) A = = =
2−(−10) −5 1 12 − 5 1 −5/12 1/12
© 2008, 2012 Zachary S Tseng D▯1 ▯ 10 7. Systems of linear equations (also known as linear systems)
A system of linear (algebraic) equations, Ax = b, could have zero,
exactly one, or infinitely many solutions. (Recall that each linear
equation has a line as its graph. A solution of a linear system is a
common intersection point of all the equations’ graphs − and there are
only 3 ways a set of lines could intersect.)
If the vector b on the right▯hand side is the zero vector, then the
system is called homogeneous. A homogeneous linear system always
has a solution, namely the all▯zero solution (that is, the origin). This
solution is called the trivial solution of the system. Therefore, a
homogeneous linear system Ax = 0 could have either exactly one
solution, or infinitely many solutions. There is no other possibility,
since it always has, at least, the trivial solution. If such a system has n
equations and exactly the same number of unknowns,then the number
of solution(s) the system has can be determined, without having to
solve the system, by the determinant of its coefficient matrix:
Theorem: If A is an n × n matrix, then the homogeneous linear
system Ax = 0 has exactly one solution (the trivial solution) if and
only if A is invertible (that is, it has a nonzero determinant). It
will have infinitely many solutions (the trivial solution, plus
infinitely many nonzero solutions) if A is not invertible
(equivalently, has zero determinant).
© 2008, 2012 Zachary S Tseng D▯1 ▯ 11 8. Eigenvalues and Eigenvectors
Given a square matrix A, suppose there are a constant r and a nonzero
vector x such that
Ax = rx,
then r is called an Eigenvalue of A, and x is an Eigenvector of A
corresponding to r.
Do eigenvalues/vectors always exist for any given square matrix?
The answer is yes. How do we find them, then?
Rewrite the above equation, we get Ax − r = 0. The next step would
be to factor out x. But doing so would give the expression
(A − r)x = 0.
Notice that it requires us to subtract a number from an n × n matrix.
That’s an undefined operation. Hence, we need to further refined it by
rewriting the term rx = rIx, and then factoring out x, obtaining
(A − rI)x = 0.
This is an n × n system of homogeneous linear (algebraic) equations,
where the coefficient matrix is (A − rI). We are looking for a nonzero
solution x of this system. Hence, by the theorem we have just seen,
the necessary and sufficient condition for the existence of such a
nonzero solution, which will become an eigenvectorof A, is that the
coefficient matrix (A − rI) must have zero determinant. Set its
determinant to zero and what we get is a degree n polynomial
equation in terms of r. The case of a 2 × 2 matrix is as follow:
a b 1 0 a − r b
A − rI = − r = .
c d 0 1 c d − r
© 2008, 2012 Zachary S Tseng D▯1 ▯ 12 Its determinant, set to 0, yields the equation
a r b
det = (a−r)(d −r)−bc = r −(a+d)r +(ad −bc) = 0
c d r
It is a degree 2 polynomial equation of r, as you can see.
This polynomial on the left is called the characteristic polynomial of
the (original) matrix A, and the equation is theteristic equation
of A. The root(s) of the characteristic polynomial are the eigenvalues
of A. Since any degree n polynomial always has n roots (real and/or
complex; not necessarily distinct), any n × n matrix always has at least
one, and up to n different eigenvalues.
Once we have found the eigenvalue(s) of the given matrix, we put
each specific eigenvalue back into the linear system (A − rI)x = 0 to
find the corresponding eigenvectors.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 13 2 3
Examples: A =
4 3
2 3 − r 1 0 = 2 − 3
A − rI =4 3 0 1 4 3−r .
Its characteristic equation is
2−r 3 2
det = (2−r)(3−r)−12 = r −5r −6 = (r +1)(r −6) = 0
4 3 −
The eigenvalues are, therefore, r = −1 and 6.
Next, we will substitute each of the 2 eigenvalues into the matrix
equation (A − rI) x = 0.
For r = −1, the system of linear equations is
2+1 3 3 3
(A − rI) x = (A +I) x = x = x = .
4 3+1 4 4
Notice that the matrix equation represents a degenerated system of 2
linear equations. Both equations are constant multiples of the
equation 1 + 2 = 0. There is now only 1 equation for the 2
unknowns, therefore, there are infinitely many possible solutions.
This is always the case when solving for eigenvectors. Necessarily,
there are infinitely many eigenvectors corresponding to each
eigenvalue.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 14 Solving the equation1x +2x = 0, we get the relat2on x 1 −x . Hence,
the eigenvectors corresponding to r = −1 are all nonzero multiples of
1
k1= .
−1
Similarly, for r = 6, the system of equations is
2−6 3 − 4 3
(A − rI) x = (A − 6I) x = x = x = .
4 3−6 4 −3
Both equations in this second linear system are equivalent to
4x1− 3x 2 0. Its solutions are given by the relatio1 4x =23x .
Hence, the eigenvectors corresponding to r = 6 are all nonzero
multiples of
k2= 4 .
Note: Every nonzero multiple of an eigenvector is also an eigenvector.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 15 Two short▯cuts to find eigenvalues:
1. If A is a diagonal or triangular matrix, that is, if it has the form
a 0 a b a 0
0 d , or 0 d , or c d .
Then the eigenvalues are just the main diagonal entries, r = a and d in all 3
examples above.
2. If A is any 2 × 2 matrix, then its characteristic equation is
a −r b
det = r2−( a d r( ad −bc ) = 0
c d −r
If you are familiar with terminology of linear algebra, the characteristic
equation can be memorized rather easily as
2
r − Trace(A)r + det(A) = 0.
Note: For any square matrix A, Trace(A) = [sum of all entries on the main
diagonal (running from top▯left to bottom▯right)]. For a 2 × 2 matrix A,
Trace(A) = a + d.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 16 A short▯cut to find eigenvectors (of a 2× 2 matrix):
Similarly, there is a trick that enables us to find the eigenvectors of any 2 × 2
matrix without having to go through the whole process of solving systems of
linear equations. This short▯cut is especially handy when the eigenvalues
are complex numbers, since it avoids the need to solve the linear equations
which will have complex number coefficients. (Warning: This method does
not work for any matrix of size larger than 22.)
We first find the eigenvalue(s) and then write down, for each eigenvalue, the
matrix (A − rI) as usual. Then we take any row of (A − rI) that is not
consisted of entirely zero entries, say it is the row vector (α , β). We put a
minus sign in front of one of the entries, for example, (α , −β). Then an
engenvector of the matrix A is found by switching the two entries in the
above vector, that is, k = (−β , α).
2 3
Example: Previously, we have seenA = 4 3 .
The characteristic equation is
r − Trace(A)r + det(A) = r − 5r − 6 = (r + 1)(r − 6) =0,
3 3
which has roots r = −1 and 6. For r = −1, the matrix (A − rI) i4 4 .
Take the first row, (3, 3), which is a non▯zero vector; put a minus sign to the
first entry to get (−3, 3); then switch the entry, we now ha1e k = (3, −3). It
is indeed an eigenvector, since it is a nonzero constant multiple of the vector
we found earlier.
On very rare occasions, both rows of the matrix (A − rI) have all zero
entries. If so, the above algorithm will not be able to find an eigenvector.
Instead, under this circumstance any non▯zero vector will be an eigenvector.
© 2008, 2012 Zachary S Tseng D▯1 ▯ 17 Exercises D▯1.2:
−5 −1 2 0
Let C = 7 3 and D = −2 −1 .
1. Compute: (i) C + 2D and (ii) 3C – 5D.
2. Compute: (i) CD and (ii) DC.
3. Compute: (i) det(C), (ii) det(D), (iii) det(CD), (iv) det(DC).
−1 −1 −1
4. Compute: (i) C , (ii) D , (iii) (CD) .
5. Find the eigenvalues and their corresponding eigenvectors of C and D.
Answers D▯1.2:
−1 −1 −25 −3
1. (i) 3 1 , (ii) 31 14
−8 1 −10 −2
2. (i) 8 −3 , (ii) 3 −1
3. (i) −8, (ii) −2, (iii) 16, (iv) 16
−3/8 −1/8 1/2 0 −3/16 −1/16
4. (i) , (ii) , (iii)
7/8 5/8 −1 −1 −1/2 −1/2
s s
5. (i) r1= 2, k1= ; 2 = −4, k2= ; s = any nonzero number
−7s − s
s
(ii) 1 = 2, k1= ; r2= −1, k2= ; s = any nonzero number
−2s/3
© 2008, 2012 Zachary S Tseng D▯1 ▯ 18 Solution of 2 × ×× 2 systems of first order linear equations
Consider a system of 2 simultaneous first order linear equations
x 1 = ax 1 bx 2
x ′ = cx + dx
2 1 2
It has the alternate matrix▯vector representation
a b
x′ = x.
c d
Or, in shorthand x′ = Ax, if A is already known from context.
We know that the above system is equivalent to a second order
homogeneous linear differential equation. As a result, we know that the
general solution contains two linearly independent parts. As well, the
solution will be consisted of some type of exponential functions. Therefore,
assume that x = ke is a solution of the system, where k is a vector of
coefficients (of 1 and x2). Substitute x and x′ = rke into the equation
x′ = Ax, and we have
rt rt
rke = Ake .
Since e is never zero, we can always divide both sides bye and get
rk = Ak.
We see that this new equation is exactly the relation that defines eigenvalues
and eigenvectors of the coefficient matrix A. In other words, in order for a
rt
function x = ke to satisfy our system of differential equations, the number r
must be an eigenvalue of A, and the vector k must be an eigenvector of A
corresponding to r. Just like the solution of a second order homogeneous
linear equation, there are three possibilities, depending on the number of
distinct, and the type of, eigenvalues the coefficient matrix A has.
© 2008, 2012 Zachary S Ts

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