Joseph m mahaffy h mahaffymathsdsuedu i lecture notes

Joseph M. Mahaffy, h [email protected] i
Lecture Notes – Second Order Linear Equations
— (9/32)
Introduction
Theory for 2
nd
Order DEs
Linear Constant Coefficient DEs
Existence and Uniqueness
Linear Operators and Superposition
Wronskian and Fundamental Set of Solutions
Wronskian
Wronskian: Consider the linear homogeneous 2 nd order DE
L [ y ] = y
00
+ p ( t ) y
0
+ q ( t ) y = 0 .
with p ( t ) and q ( t ) continuous on an interval I
Let y
1
and y
2
be solutions satisfying L [ y
i
] = 0 for i = 1 , 2 and define
the Wronskian by
W [ y
1
, y
2
]( t ) =
y
1
( t ) y
2
( t )
y 0
1
( t ) y 0
2
( t )
= y
1
( t ) y
0
2
( t ) - y
0
1
( t ) y
2
( t ) .
If W [ y
1
, y
2
]( t ) 6 = 0 on I , then the general solution of L [ y ] = 0
satisfies
y ( t ) = c
1
y
1
( t ) + c
2
y
2
( t ) .
Joseph M. Mahaffy, h [email protected] i
Lecture Notes – Second Order Linear Equations
— (10/32)
Introduction
Theory for 2
nd
Order DEs
Linear Constant Coefficient DEs
Existence and Uniqueness
Linear Operators and Superposition
Wronskian and Fundamental Set of Solutions
Fundamental Set of Solutions
Theorem
Let y
1
and y
2
be two solutions of
y
00
+ p ( t ) y
0
+ q ( t ) y = 0 ,
and assume the Wronskian, W [ y
1
, y
2
]( t ) 6 = 0 on I . Then y
1
and y
2
form a fundamental set of solutions , and the general solution is
given by
y ( t ) = c
1
y
1
( t ) + c
2
y
2
( t ) .
where c
1
and c
2
are arbitrary constants. If there are given initial
conditions, y ( t
0
) = y
0
and y 0 ( t
0
) = y
1
for some t
0
∈ I , then these
conditions determine c
1
and c
2
uniquely.
Joseph M. Mahaffy, h [email protected] i
Lecture Notes – Second Order Linear Equations
— (11/32)
Introduction
Theory for 2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Homogeneous Equations 1
Homogeneous Equation: The general 2
nd
order constant
coefficient homogeneous differential equation is written:
ay
00
+ by
0
+ cy = 0
This can be written as a system of 1
st
order differential
equations
˙ x = Ax =
0 1
- c/a - b/a
x ,
where
x =
x
1
x
2
=
y
y
0
This has a the general solution
x = c
1
y
1
y
0
1
+ c
2
y
2
y
0
2
Joseph M. Mahaffy, h [email protected] i
Lecture Notes – Second Order Linear Equations
— (12/32)

Introduction
Theory for 2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Homogeneous Equations 2
Characteristic Equation: Obtain characteristic equation by
solving
det | A - λ I | =
- λ 1
- c/a - b/a - λ
=
1
a
(
aλ 2 + bλ + c
)
= 0
Find eigenvectors by solving
( A - λ I ) v =
- λ 1
- c/a - b/a - λ
v
1
v
2
=
0
0
If λ is an eigenvalue, then it follows the corresponding eigenvector is
v =
1
λ
Then a solution is given by
x = e
λt
v =
e λt
λe
λt
=
y ( t )
y
0
( t )
Joseph M. Mahaffy, h [email protected] i
Lecture Notes – Second Order Linear Equations
— (13/32)
Introduction
Theory for 2
nd
Order DEs
Linear Constant Coefficient DEs
Homogeneous Equations
Method of Undetermined Coefficients
Forced Vibrations
Homogeneous Equations 3
Theorem
Let λ
1
and λ
2
be the roots of the characteristic equation
aλ
2
+ bλ + c = 0 .