ode ode Basic Concepts and Theorems The n th order linear ODE takes - - PowerPoint PPT Presentation
ode ode Basic Concepts and Theorems The n th order linear ODE takes - - PowerPoint PPT Presentation
So Solution tion of of hi high gher er or orde der r ode ode Basic Concepts and Theorems The n th order linear ODE takes the form: n n 1 d y d y dy a a ... a a y f x ( ) n n 1 1
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Basic Concepts and Theorems
The nth order linear ODE takes the form:
1 1 1 1
... ( )
n n n n n n
d y d y dy a a a a y f x dx dy dx
(1) y and all its derivatives are of the first degree
1
, ,...,
n n
a a a
(2) All the coefficients are function of the independent variable x only or constants.
Definition:
The D operator is defined to be:
d D dx
dy Dy dx
2 2 2
d D dx
2 2 2
d y D y dx
n n n
d D dx
n n n
d y D y dx
are n independent solutions of the homogenous DE
1 1 2 2
...
n n
y x c y c y c y
1 2
, ,..., n y y y
then the general solution is
Theorem
Complementary and particular solutions
The complementary (or homogenous) solution:
c
y
p
y
G
y
G c p
y y y
The particular solution: is any solution satisfy the non-homogenous case is the solution of the homogenous case The general solution:
Second Order Linear Homogenous DE with Constant Coefficients
1
y a y a y
x
y e
x
y e
2 x
y e
2 1
( )
x
a a e
2 1
( ) a a
This equation called the characteristic equation (c/e) (or auxiliary equation) of the DE.
The solution is:
x x
e c e c y
2 1
2 1
6 5 y y y
6 5
2
3 2
1 2
2, 3
x x
e c e c y
3 2 2 1
The solution is:
1 2
( cos sin )
x
y e c x c x
4 13 y y y
2
4 13
1,2
2 3i
2 1 2
( cos3 sin3 )
x
y e c x c x
The solution is:
1 2 x x
y c e c xe
4 4 y y y
2
4 4
2
2
1 2
2
2 2 1 2 x x
y c e c xe
Example
2 y y y
2
1
1 2
1
1 2 x x
y c e c xe
Solution
2
2 1
Example 4 5 (0) 1 (0) y y y y y
2
4 5
5 1
1 2
5, 1
5 1 2 x x
y c e c e Solution
1 2
(0) 1 1 y c c
1 2
(0) 5 y c c
1 2
1 5 , 6 6 c c
5
1 5 6 6
x x
y e e
Example
9 y y
2
9
1 2
3i
1 2
cos3 sin3 y c x c x Solution
Higher Orders Homogenous Linear DE with Constant Coefficients
1 1 1
...
n n n n
a y a y a y a y
1 1 1
...
n n n n
a a a a
1 2
, ,..., n
(3) According to the nature of these roots we write the linearly independent solutions. (1) Set the c/e: (2) Find the roots:
Example 2 3 y y y
Solution
3 2
2 3
2
2 3
3 1
1 2 3
0, 1, 3
3 1 2 3 x x x
y c e c e c e
3 1 2 3 x x
y c c e c e
Example 4 4 y y y y
Solution
3 2
4 4
2
1
4 1
2
1 4
1 2,3
1, 2i
1 2 3
cos2 sin2
x
y c e c x c x
Example
4 2
2 1 D D y
Solution
4 2
2 1
2 2
( 1)
1,2 3,4
i
1 2 3 4
( cos sin ) ( cos sin ) y c x c x x c x c x
Example
4 2
8 16 D D y
Solution
4 2
8 16
2 2
( 4)
2 2
( 2) ( 2)
1,2 3,4
2, 2
2 2 2 2 1 2 3 4
( ) ( )
x x x x
y c e c xe c e c xe