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Unformatted text preview: Examples Example: Consider the IVP 2 x 2 y 00 + 3 xy 3 y = 5 x y (1) = 1 ,y (1) = 1 . ( NH ) Strategy: find a FS for the associated ho mogeneous SOLODE 2 x 2 y 00 + 3 xy 3 y = 0 , ( H ) then, using Variation of Parameters, find a particular solution Y ( x ) of ( NH ) , which will give us the general solution y ( x ) = Y ( x ) + c 1 y 1 ( x ) + c 2 y 2 ( x ) , ( G ) then accommodate the Initial Conditions. See that ( H ) has the solution y 1 ( x ) = x . Panic! I have forgotten how Reduction of Order works. Actually, I do remember that it uses Abels theorem: there is a second solution y 2 such that W [ y 1 ,y 2 ]( x ) = e R p ( x ) dx . Now y 1 y 2 y 2 y 1 = e R 3 / 2 x dx = x 3 / 2 = x 3 / 2 = y 1 y 2 y 2 y 1 y 2 1 = x 3 / 2 x 2 = x 7 / 2 = y 2 y 1 = x 7 / 2 = y 2 y 1 = Z x 7 / 2 dx = 2 5 x 5 / 2 2 = y 2 ( x ) = 5 2 x 3 / 2 Thus { x,x 3 / 2 } is a FS for ( H ) . It has Wronskian W [ y 1 ,y 2 ] = 5 2 x 3 / 2 . Now find a particular solution Y of ( NH ) . Panic! I have forgotten how Variation of Parameters works. Actually, I remember that it means looking for a particular so lution of the form Y ( x ) = v 1 ( x ) y 1 ( x ) + v 2 ( x ) y 2 ( x ) with variable parameters v 1 ,v 2 . So I take some scratch paper and derive it right quick: 3 Y 00 = v 1 y 00 1 + v 2 y 00 2 + v 1 y 1 + v 2 y 2 p Y = v 1 p y 1 + v 2 p y 2 + p [ v 1 y 1 + v 2 y 2  {z } =0 ] q Y = v 1 q y 1 + v 2 q y 2 L [ Y ] = v 1 L [ y 1 ] + v 2 L [ y 2 ]  {z } =0 + v 1 y 1 + v 2 y 2  {z } = g results in the two linear equations...
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This note was uploaded on 09/07/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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