285lect18 - 1 1.1 Lecture 18 Non-homogeneous linear ODE of order n Let L = Dn p1(x Dn 1 p2(x Dn 2 pn 1(x D pn(x D0 Consider non-homogeneous ODE y(n

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1 Lecture 18 1.1 Non-homogeneous linear ODE of order n Let L = D n + p 1 ( x ) D n 1 + p 2 ( x ) D n 2 + ::: + p n 1 ( x ) D + p n ( x ) D 0 : Consider non-homogeneous ODE y ( n ) + p 1 ( x ) y ( n 1) + p 2 ( x ) y ( n 2) + ::: + p n ( x ) y ( x ) = f ( x ) , or L [ y ] = f: The complementary homogeneous equation is y ( n ) + p 1 ( x ) y ( n 1) + p 2 ( x ) y ( n 2) + ::: + p n ( x ) y ( x ) = 0 , or L [ y ] = 0 : Solutions of complementary equation are called complementary solutions y c ( x ) . So, L [ y c ] = 0 . Let y p ( x ) homogeneous equation L [ y ] = f . Hence, L [ y p ] = f . Then y ( x ) = y c ( x ) + y p ( x ) is a solution of the non-homogeneous equation. Indeed, L [ y c + y p ] = L [ y c ] + L [ y p ] = 0 + f ( x ) = f ( x ) : Notice, that if y c is a general solution of the complimentary equation, then, there are n constants in the formula for y c . Thus, y ( x ) = y c ( x ) + y p ( x ) contains n constants C 1 ; C 2 ; :::; C n . It can be showed that by selecting the constants C 1 ; C 2 ; :::; C n , we can solve a given initial value problem. Example 1
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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285lect18 - 1 1.1 Lecture 18 Non-homogeneous linear ODE of order n Let L = Dn p1(x Dn 1 p2(x Dn 2 pn 1(x D pn(x D0 Consider non-homogeneous ODE y(n

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