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

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

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

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