ode ode Basic Concepts and Theorems The n th order linear ODE takes - PowerPoint PPT Presentation

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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 0 n n 1 dx dy dx (1) y and all its derivatives are of the first degree a , a ,..., a (2) All the coefficients are function of the  n n 1 0 independent variable x only or constants.

Definition: The D operator is defined to be: dy d   Dy D dx dx 2 2 d d y 2  2  D D y 2 2 dx dx n n d d y   n n D D y n n dx dx Theorem y y , ,..., n y are n independent solutions of the homogenous DE 1 2       y x c y c y ... c y then the general solution is 1 1 2 2 n n

Complementary and particular solutions The complementary (or homogenous) solution: y is the solution of the homogenous case c The particular solution: y is any solution satisfy the non-homogenous case p y The general solution: G   y y y G c p

Second Order Linear Homogenous DE with Constant Coefficients e   x      y y a y a y 0 1 0 e  e     2 x    x y y a e        2     2 x ( a a ) 0 ( a ) 0 1 0 1 0 This equation called the characteristic equation (c/e) (or auxiliary equation) of the DE.

    x x y c e c e 1 2 The solution is: 1 2       5 6 0 y y y      2 5 6 0     2 x 3 x y c e c e 1 2         2 3 0        2, 3 1 2

The solution is:      x y e ( c cos x c sin x ) 1 2      y 4 y 13 y 0      2 4 13 0    2 x y e ( c cos3 x c sin3 ) x 1 2     2 3 i 1,2

    x x y c e c xe The solution is: 1 2      y 4 y 4 y 0  2     4 4 0   2 x 2 x y c e c xe 1 2   2    2 0      2 1 2

Example      y 2 y y 0 Solution  2     2 1 0     x x   2 y c e c xe    1 2 1 0       1 1 2

Example         y 4 y 5 y 0 y (0) 1 y (0) 0 Solution      2     4 5 0 y (0) 1 c c 1 1 2              y (0) 0 5 c c 0 5 1 0 1 2       1 5   5, 1 c , c 1 2 1 2 6 6 c e   5  x x 1 5 e  y c e   5 x x y e 1 2 6 6

Example    y 9 y 0 Solution    2 9 0       3 i 1 2   y c cos3 x c sin3 x 1 2

Higher Orders Homogenous Linear DE with Constant Coefficients      n n 1       a y a y ... a y a y 0  n n 1 1 0     1      n n (1) Set the c/e: a a ... a a 0  n n 1 1 0    , ,..., n (2) Find the roots: 1 2 (3) According to the nature of these roots we write the linearly independent solutions.

Example       y 2 y 3 y 0 Solution         3   2    0, 1, 3 2 3 0 1 2 3       0 x x 3 x       2 y c e c e c e 2 3 0 1 2 3        3 x x       y c c e c e 3 1 0 1 2 3

Example        y y 4 y 4 y 0 Solution          2  3   2     1 4 0 4 4 0              2    1, 2 i 1 4 1 0 1 2,3  x   y c e c cos2 x c sin2 x 1 2 3

Example   4  2   D 2 D 1 y 0 Solution      4 2 2 1 0      i 1,2 3,4   2 2  ( 1) 0     y ( c cos x c sin ) x x c ( cos x c sin ) x 1 2 3 4

Example   4  2   D 8 D 16 y 0 Solution      4 2   2   2  8 16 0 ( 2) ( 2) 0         2 2 2, 2 ( 4) 0 1,2 3,4       2 x 2 x 2 x 2 x y ( c e c xe ) ( c e c xe ) 1 2 3 4