PERTEMUAN XIII
SISTEM PERSAMAAN SIMULTAN
.
Systems of first-order linear differential equations
Two equations in two variables
Consider the system of linear differential equations (with constant coefficients)
x(t) = ax(t) + by(t)
y(t) = cx(t) + dy(t).
We can solve this system using the techniques from the previous section, as
follows. First isolate y(t) in the first equation, to give
y(t) = x(t)/b  ax(t)/b.
Now differentiate this equation, to give
y(t) = x(t)/b  ax(t)/b.
Why is this step helpful? Because we can now substitute for y(t) and y(t) in the
second of the two equations in our system to yield
x(t)/b  ax(t)/b = cx(t) + d[x(t)/b  ax(t)/b],
which we can write as
x(t)  (a + d)x(t) + (ad  b)x(t) = 0,
an equation we know how to solve!
Having solved this linear second-order differential equation in x(t), we can go
back to the expression for y(t) in terms of x(t) and x(t) to obtain a solution for
y(t).
(We could alternatively have started by isolating x(t) in the second equation and
creating a second-order equation in y(t).)
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Example
Consider the system of equations
x(t) = 2x(t) + y(t)
y(t) = 4x(t)  3y(t).
Isolating y(t) in the first equation we have y(t) = x(t)  2x(t), so that
y(t) = x(t)  2x(t). Substituting these expressions into the second
equation we get
x(t)  2x(t) = 4x(t)  3x(t) + 6x(t),
or
x(t) + x(t)  2x(t) = 0.
We have seen that the general solution of this equation is
x(t) = Aet + Be2t.
Using the expression y(t) = x(t)  2x(t) we get
y(t) = Aet  2Be2t  2Aet  2Be2t,
or
y(t) = Aet  4Be2t.
General linear systems
We may write the system
x(t) = ax(t) + by(t)
y(t) = cx(t) + dy(t),
studied above, as
x(t)
y(t)
=
a b
c d
x(t)
y(t)
.
I give only one example, which shows how the trigonometric functions may
emerge in the solution of a system of two simultaneous linear equations,
which, as we saw above, is equivalent to a second-order equation.
Example
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Consider the system
x1(t)
0 b
x1(t)
=
.
x2(t)
b 0
x2(t)
We need to find Ak for each value of k, where
0 b
.
b 0
You should be able to convince yourself (by computing A2, A3, and A4)
that if k is odd we have
A=
k
A =
0
(1)(k+1)/2bk
0
(1)(k1)/2bk
whereas if k is even we have
,
0
(1)k/2bk
.
0
(1)k/2bk
Given these results, we have
Ak =
1  b2/2! + b4/4!  b6/6! + ...
b + b3/3!  b5/5! + ...
e =
b  b3/3! + b5/5!  ...
1  b2/2! + b4/4! + b6/6! + ...
Now, you should recall that
sin b = b  b3/3! + b5/5!  ...
and
cos b = 1  b2/2! + b4/4!  b6/6! + ...
Thus
A
.
cos b sin b
.
sin b cos b
We conclude that the solution of the system of equations given the initial
conditions x1(0) = C1 and x2(0) = C2 is
x1(t) = C1cos bt  C2sin bt
x2(t) = C1sin bt + C2cos bt.
You are asked, in an exercise, to verify this solution by using the
technique discussed at the start of this section to convert the twoequation system to a single second-order linear differential equation.
eA =
Exercises on systems of first-order linear differential
equations
1. Find the general solution of the system of equations
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x(t) = 4y(t)
y(t) = x(t) + 4y(t).
Solutions
1. Isolating y(t) in the first equation we have y(t) = x(t)/4, so that y(t) =
x(t)/4. Substituting these expressions into the second equation we get
x(t)/4 = x(t) + x(t),
or
x(t)  4x(t) + 4x(t) = 0.
The characteristic equation is (r  2)2, which has the repeated root r = 2.
Thus the general solution of this equation is
x(t) = (A + Bt)e2t.
Given y(t) = x(t)/4, we thus have
y(t) = (1/2)(A + Bt)e2t + (1/4)Be2t = [(1/4)B + (1/2)A + (1/2)Bt]e2t.
========================================
2. Find the general solution of the following system of equations and the
particular solution with x(0) = 1 and y(0) = 0.
x(t) = x(t)  5y(t)
y(t) = 2x(t)  5y(t).
Solutions
x(t) = e2t[(3A  B)cos t  (A + 3B)sin t]
y(t) = e2t[2Acos t  2Bsin t].
For the given initial conditions we have A = 0 and B = 1.
3. Find the solution of the system
x(t) = by(t)
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y(t) = bx(t)
given the initial conditions x(0) = C1 and y(0) = C2.
Solutions
We have x(t) = by(t) = b2x(t), or
x(t) + b2x(t) = 0.
The characteristic equation is r2 + b2 = 0, which has complex roots.
Thus, given the general form of the solution to such an equation, the
solution of our equation is
x(t) = C1cos bt  C2sin bt.
Using y(t) = x(t)/b, we have y(t) = C1sin bt + C2cos bt. (Remember that
the derivative of sin bt is bcos bt, and the derivative of cos bt is
bsin bt.)
Example:
Following the method just proposed, we obtain
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After applying the initial conditions, we find A = 1 and B = -3/2 .
After applying the initial conditions, we find A=1 and B=-3/2 .
Example 1.
Solve for t ≥ 0 the first-order simultaneous differential equation
x′ + y′ + 5x + 3y = e-t
2x′ + y′ + x + y = 3
initial conditions x = 2 and y = 1 at t = 0
Jawab :
x′ + y′ + 5x + 3y = e-t  ( D + 5) x + ( D + 3 ) y = e -t
2x′ + y′ + x + y = 3  ( 2D + 1 ) x + ( D + 1 )y = 3
result
| x(t)=25*e^(t)/3-11*e^(-2*t)-9/2 |
||
| y(t)=-25*e^(t)/2+1*e^(-t)/2+11*e^(-2*t)/2+15/2 |
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Example:
Then the characteristic equation and the eigenvalues are:
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The system is unstable.
Method 2: By Operator Matrices
Step 1: Replace the derivatives by differential operators, D =
d/dt,
Step 2: Extract the matrix operator,
Step 2: Use Cramer’s rule,
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or,
which is the same result as Method 1 when D is replaced back by
d/dt.
Generalization:
The characteristic equation for a set of n linear first order ODEs
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can be obtained as
Example:
For the system given by
Content
ERIMA KASIH
http://www.chass.utoronto.ca/~osborne/MathTutorial/SDE.HTM
http://www.chass.utoronto.ca/~osborne/MathTutorial/SDE.HTM#complexroots
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