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Unformatted text preview: Ch 3.2: Fundamental Solutions of Linear Homogeneous Equations Let p , q be continuous functions on an interval I = ( , ), which could be infinite. For any function y that is twice differentiable on I , define the differential operator L by Note that L [ y ] is a function on I , with output value For example, [ ] y q y p y y L + + = [ ] ) ( ) ( ) ( ) ( ) ( ) ( t y t q t y t p t y t y L + + = ( 29 [ ] ) sin( 2 ) cos( ) sin( ) ( 2 , ), sin( ) ( , ) ( , ) ( 2 2 2 2 t e t t t t y L I t t y e t q t t p t t + + = = = = = Differential Operator Notation In this section we will discuss the second order linear homogeneous equation L [ y ]( t ) = 0, along with initial conditions as indicated below: We would like to know if there are solutions to this initial value problem, and if so, are they unique. Also, we would like to know what can be said about the form and structure of solutions that might be helpful in finding solutions to particular problems. These questions are addressed in the theorems of this section. [ ] 1 ) ( , ) ( ) ( ) ( y t y y t y y t q y t p y y L = = = + + = Theorem 3.2.1 Consider the initial value problem where p , q , and g are continuous on an open interval I that contains t . Then there exists a unique solution y = ( t ) on I. Note: While this theorem says that a solution to the initial value problem above exists, it is often not possible to write down a useful expression for the solution. This is a major difference between first and second order linear equations. ) ( , ) ( ) ( ) ( ) ( y t y y t y t g y t q y t p y = = = + + Example 1 Consider the second order linear initial value problem In Section 3.1, we showed that this initial value problem had the following solution: Note that p ( t ) = 0, q ( t ) = 1, g ( t ) = 0 are each continuous on ( , ), and the solution y is defined and twice differentiable on ( , ). t t e e t y + = 2 ) ( ( 29 ( 29 1 , 3 , = = = y y y y 1 ) ( , ) ( ) ( ) ( ) ( y t y y t y t g y t q y t p y = = = + + Example 2 Consider the second order linear initial value problem where p , q are continuous on an open interval I containing t . In light of the initial conditions, note that y = 0 is a solution to this homogeneous initial value problem. Since the hypotheses of Theorem 3.2.1 are satisfied, it follows that y = 0 is the only solution of this problem....
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This note was uploaded on 08/13/2009 for the course DIFF 2343632 taught by Professor Csar during the Fall '09 term at Middle East Technical University.
 Fall '09
 Csar

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