HOMOGENEOUS LINEAR SYSTEM WITH CONSTANT COEFFICENTS

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HOMOGENEOUS LINEAR SYSTEM WITH CONSTANT
COEFFICENTS
TWO EQUATIONS IN TWO UNKNOWN FUNCTIONS
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We shall be concerned with the homogeneous linear
system
where the coefficients a1, a2, b1, and b2 are
real constants. We seek solution of this system.
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If we recall that the exponential function has
the property that its derivatives are constant
multiples of the function itself then, let us
therefore attempt to determine a solution of the
form
where A, B, and m are constants. If we
substitute (2) into (1), we obtain
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These equations lead at once to the system
in the unknowns A and B. This system obviously
has the trivial solution A B 0. But this
would only lead to the trivial solution x y
0 of system (1).
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Thus we seek nontrivial solutions of the system
(1).
A necessary and sufficient condition that this
system has a nontrivial solution is that the
determinant of the coefficients
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Expanding this determinant we get the quadratic
equation
in the unknown m.
This equation is called the Characteristic
equation (or Auxiliary equation) associated with
the system (1). Its roots m1 and m2 are called
the Characteristic roots.
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Three cases must now be considered

• The roots m1 and m2 are real and distinct.
• The roots m1 and m2 real and equal.
• The roots m1 and m2 are complex (and hence are
  conjugates of each other) .

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Case I The roots of the Characteristic equations
(5) are real and distinct say m1,m2
Let m m1. We note then that the two
equations in (3) are LD i.e. One equation will
be a scalar multiple of the other. So we
consider one of them and get one nontrivial
solution A1, B1 of this algebraic system. Hence
we get the nontrivial solution
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Similarly corresponding to the root m m2, we
get a second solution
10
We can show that these two solutions are LI and
hence the general solution is
c1, c2 arbitrary constants.
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Example
Solution
The Characteristic equation is
12
Expanding this we obtain
or
Solving this we find the roots m1 4, m2 -1.
Corresponding to m1 4, A, B are got from
13
i.e.
These are same as
Hence one notrivial solution is A 2, B 3 and
the corresponding solution of the system is
14
Corresponding to m2 -1, A, B are got from
These are same as
Hence one notrivial solution is A 1, B -1
and the corresponding solution of the system is
15
Hence the general solution is
c1, c2 arbitrary constants.
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Example
Solution
The Characteristic equation is
17
Expanding this we obtain
Solving this we find the roots m1 3, m2 4.
Corresponding to m1 3, A, B are got from
18
i.e.
These are same as
Hence one notrivial solution is A B 1 and the
corresponding solution of the system is
19
Corresponding to m2 4, A, B are got from
These are same as
Hence one notrivial solution is A 3, B 2 and
the corresponding solution of the system is
20
Hence the general solution is
c1, c2 arbitrary constants.
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Case II. The roots of the Characteristic equation
are real and equal
If the roots of the charactrisic equation (1)
are real and equal, i.e. m1 m2. then
corresponding to the root m m1, we get a
solution
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A second LI solution will be of the form
The constants A2, A3, B2, and B3 are got by
substituting for x, y in the given system of
differential equations. We explain by an example.
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Example Solve the system
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The characteristic equation is
i.e.
Roots are m 3, 3 (Real and equal)
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Corresponding to m 3, A, B are got from the
equation
i.e.
A nontrivial solution is A 2, B -1
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Corresponding solution of the system is
We assume a second LI solution as
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Substituting in the given system, we get
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Simplifying, we get
Equating the coefficients of t and constant term
to zero we get
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From the first two equations, we get a nontrivial
solution for A3, B3 as A3 2, B3 -1
Using these values in the second system, we get
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Hence we get a nontrivial solution for A2, B2 as
A2 1, B2 0
Hence a second LI solution is
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Hence the general solution of the given system is
c1, c2 arbitrary constants
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Case III. The roots of the Characteristic
equation are complex conjugates
If the roots of the charactrisic equation are the
conjugate complex numbers
then we obtain two complex solutions
and
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In this case, the real and imaginary parts of the
first solution form two LI real solutions of the
system and the general solution is the real
linear combination of them. Again we explain by
means of an example.
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Example Solve the system
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The characteristic equation is
i.e.
Roots are
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Corresponding to
A, B are got from the equation
A nontrivial solution is
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Corresponding solution of the system is
Their real parts are
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Their imaginary parts are
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Hence the general solution is
c1, c2 arbitrary constants
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Example
We assume a solution of the form
Substituting (12) into (11) we obtain
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and this leads at once to the algebric system
in the unknown m. For nontrivial solutions of
this system we must have
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Expanding this, we obtain the characteristic
equation
The roots of this equation are the conjugate
complex numbers
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Setting m 23i in (13), we obtain
A simple nontrivial solution of this system is A
2, B -13i. Using these values we obtain the
complex solution
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Using Eulers formula this takes the form
Since both the real and imaginary parts of this
solution of system (11) are themselves solution
of (11), we thus obtain the two real solutions
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and
46
Finally, since the two solutions are linearly
independent we may write the general solution of
the system (11) in the form
where c1 and c2 are arbitrary constants.
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Example
We assume a solution of the form
Substituting (16) into (15) we obtain
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and this leads at once to the algebric system
in the unknown m. For nontrivial solutions of
this system we must have
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Expanding this we obtain the characteristic
equation
Thus the characteristic equation (18) has the
real and equal roots 3, 3.
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Setting m 3 in (18), we obtain
A simple nontrivial solution of this system being
A B 1, we obtain the nontrivial solution
of the given system (15)
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Since the roots of the characteristic equation
are both equla to 3, we must seek a second
solution of the form (14) with m 3. That is,
we must determine A1, A2, B1, and B2 ( with A1
and B1 not both zero) such that
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is a solution of the system (15). Substitutiong
(20) into (15), we obtain
These equations reduce at once to
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Equating the coefficients of t and the constant
terms to zero, we get
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Thus in order for (21) to be a solution of the
system (15), the constsnts A2, A3, B2, and B3
must be chosen to satisfy the above equations.
From the equations A2 B2 0, we see that
A2 B2. The other equations of (21) show that A2
and B2 must satisfy
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We may choose any convenient nonzero values for
A2 and B2. We Choose A2 B2 1. Then (22)
reduces to A3 B3 1, and we can choose any
convenient values for A3 and B3 that will satisfy
this equation. We choose A3 1, B3 0. We are
thus led to the solution
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By Theorem the solutions (19) and (23) are
linearly independent. We may thus write the
general solution of the system (15) of the form
where c1 and c2 are arbitrary constants.
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We note that a different choice of nonzero values
for A2 and B2 in (22) and/or a different choice
for A3 and B3 in the resulting equation for A3
and B3 will lead to a second solution which is
different from solution (23). But this different
second solution will also be linearly
independent of the basic solution (19), and hence
could serve along with (19) as one of the two
constitutent parts of a general solution.
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For example, if we choose A2 B2 2 in (22),
then (22) reduced to A3 B3 2 and if we then
chose A3 3, B3 1, we are led to the different
second solution
This is linearly independent of solution (19) and
could serve along with (19) as one of the two
constitutent parts of a general solution.
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Method of Variation of Parameters (for finding
particular solution of nonhomogeneous linear
systems)
Consider the nonhomogeneous linear system
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and the corresponding homogeneous system
If
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is the general solution of the associated
homogeneous system, it is called the
complementary function (CF) of the given
non-homogeneous system.
We take a particular solution as
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where the functions v1(t) and v2(t) are chosen
to satisfy the system
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Example Find the general solution of the system
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Example Find the general solution of the system
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Solution Corresponding homogeneous system is
Its general solution is ( i.e. the CF is)
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We take a particular solution as
where
satisfy the equations
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Solving, we get
Hence
Integrating, we get
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Solving, we get
Hence
Integrating, we get
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Hence
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Homework Do the same problem by the method of
undetermined coefficients
That is assume a particular solution as
a, b, c, d constatnts.
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Example Find the general solution of the system
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Solution Corresponding homogeneous system is
its general solution is
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Formula for particular solution is
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Solving these two equations (for v1 and v2)
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Hence a particular solution of the given
non-homogeneous system is
i.e.