notes3_2 - MA366 Sathaye Notes on Chapter 3 Sec 4-8 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA366 Sathaye Notes on Chapter 3 Sec. 4-8 1. Reduction of Order. 3.4. Note that the topic of repeated roots of the characteristic equation was already discussed earlier. Here we consider a linear order 2 equation f ( D ) y = 0 and assume that we already know one non zero solution y 1 ( t ). A second solution can then be found by solving a first order equation deduced from the given equation. Here is the procedure. Set y = vy 1 . Then we have: f ( D )( vy 1 ) = vf ( D )( y 1 ) + D ( v ) f ( D )( y 1 ) + D 2 ( v ) f 00 ( D )( y 1 ) / 2! by using our Leibnitz Rule for Polynomial Operators. Suppose f ( D ) = aD 2 + bD + c where a,b,c may be constants or other functions of t . Since f ( D )( y 1 ) = 0, the above formula simplifies as follows. We first note f ( D ) = 2 aD + b and f 00 ( D ) = 2 a . Then our equation is: f ( D )( vy 1 ) = D ( v )(2 aD ( y 1 ) + by 1 ) + D 2 ( v )(2 ay 1 ) / 2 = 0 . If we set w = D ( v ), then we have: ( ay 1 ) D ( w ) + (2 aD ( y 1 ) + by 1 ) w = 0 . This is a first order equation in w , so we have reduced the order from 2 to 1. Now, solve this first order equation by known techniques to find w and then v by solving D ( v ) = w . Naturally the solution will have the form v = c 1 v 1 + c 2 for some function v 1 and arbitrary constants c 1 ,c 2 . This gives a full solution to the original equation y = ( c 1 v 1 + c 2 ) y 1 = c 1 ( v 1 y 1 ) + c 2 ( y 1 ) . 2. Particular solutions for Non homogeneous Equations. 3.5. We learn to solve f ( D )( y ) = g ( t ) when f ( D ) is a second order operator with constant coefficients. We know that the general solution has the form y = c 1 y 1 + c 2 y 2 + Y where Y is said to be a particular solution. Often the two parts are written as y h = c 1 y 1 + c 2 y 2 where y 1 ,y 2 are fundamental solutions of f ( D )( y ) = 0 with c 1 ,c 2 arbitrary constants and y p = Y is a particular solution of the full equation. Important. We recall that by a set y 1 ,y 2 of fundamental solutions, we mean two solutions which are independent in an interval near the starting point. The independence can be checked by checking the Wronskian W ( y 1 ,y 2 ) at the starting point to be non zero....
View Full Document

This note was uploaded on 01/18/2012 for the course MATH 366 taught by Professor Edraygoins during the Fall '09 term at Purdue.

Page1 / 5

notes3_2 - MA366 Sathaye Notes on Chapter 3 Sec 4-8 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online