Highlights
Math 304
Linear Algebra
From last time:
I eigenvalues and eigenvectors
Harold P. Boas
Today:
I application of eigenvectors to systems of differential
boas@tamu.edu
equations
June 27, 2006
Linear systems of differential equations
Reminders on rst-order differential equations
Exercise 1(b), page 323
Solve the differential equation dy
dt
with the initial condition y (0) = 4.
Example.
=
3y or y 0
Find the general solution to the system
=
3y
We know a function whose derivative is 3 times the
function: namely, y (t ) = e3t , or more generally y (t ) = ce3t for
an arbitrary constant c.
Solution.
Thus the general solution to y 0
=
3y is y (t ) = ce3t .
The particular solution satisfying the initial condition y (0) = 4 is
y (t ) = 4e3t .
Solution.
Write y =
system says y0
=
Ay.
y1
y2
and A =
( 0
y = 2y
1
y0
2
2
1
=
1
+ 4y2
y1
3y2 :
4
. Then the
3
Observe that if v is an eigenvector of A with eigenvalue , then
t
y(t ) = e v is a solution of the differential equation. Since
A
haseigenvalues
1 and 2 with corresponding eigenvectors
4
1
and
, the general solution is the superposition
1
1
( y (t ) = 4c et + c e 2t
1
1
2
4
1
t
2t
y (t ) = c1 e
+ c2 e
, or
t
1
1
y2 (t ) = c1 e
c2 e 2t :
An initial condition would let you determine c1 and c2 .
Differential equations with complex eigenvalues
Euler's formula
( 0
y =y
Example: exercise 1(d), page 323
Solve
The complex exponential function is related to the trigonometric
functions via Euler's formula:
eit
=
cos(t ) + i sin(t ):
1
y20
For example, ei =
1, ei =4 = (1 + i )= 2, and
e(2+3i )t = e2t (cos(3t ) + i sin(3t )).
Example: exercise 5(b), page 324
0
y 00 = 2y + y 0 :
1
1
+ y2
2
2
1
We know how to handle systems of
rst-order differential equations, so introduce two new variables
y3 and y4 via y10 = y3 and y20 = y4 . The system becomes
Solution strategy.
y10
=
y3
=
y4
3
=
2y1 + y4
4
=
2y2 + y3
y20
y0
y0
or
y
0
0
B
0 = B0
@2
0
0
0
0 2
1
0
0
1
1
0
y = Ay with A =
1
.
1
1
(1+i )t
i
1
=
et ( sin(t ) + i cos(t ))
:
et (cos(t ) + i sin(t ))
Because the differential equation is real-valued, both the real
and imaginary parts of the complex solution are real solutions.
The general (real) solution is therefore
Higher-order systems
Solve
or
y1 + y2
t) = e
y(
( 00
y = 2y
y2
The characteristic
equation is 2 2 + 2 = 0. By the quadratic
p
formula, = 2 24 8 = 1 i. An eigenvector corresponding to
i
eigenvalue 1 + i is
. One complex-valued solution is
1
y(
p
=
1
1
0
1C
C y:
1A
0
Proceed as before: nd the eigenvalues [1 and 2], the
corresponding eigenvectors, and write the general solution
[with four arbitrary constants c1 , c2 , c3 , and c4 ].
t ) = c1 e
t
sin(t )
cos(t )
+ c2 e
t
cos(t )
:
sin(t )