1.2 Problem 9 Solved

1.2 Problem 9 Solved - a 2 + b 2 ( ) u = f x ! x , ! y ( )...

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Solve au x + bu y = f x , y ( ) were f is a given function. Write the solution in the form u x , y ( ) = a 2 + b 2 ( ) ! 1/2 fds L " + g bx ! ay ( ) . Using the coordinate method: ! x = ax + by ! y = bx " ay ! u x = au ! x + bu ! y u y = bu ! x " au ! y au x + bu y = f x , y ( ) au x + bu y = a au ! x + bu ! y ( ) + b bu ! x " au ! y ( ) = a 2 + b 2 ( ) u ! x We have: a 2 + b 2 ( ) u ! x = f x , y ( ) . After some manipulation we have: x = a ! x + b ! y a 2 + b 2 ! a 2 + b 2 ( ) u ! x = f x ! x , ! y ( ) , y ! x , ! y ( ) ( ) y = b ! x " a ! y a 2 + b 2 Integrating with respect to ! x gives a line integral directed along the characteristic curve, starting at the y-axis and ending at a point x,y.
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Unformatted text preview: a 2 + b 2 ( ) u = f x ! x , ! y ( ) , y ! x , ! y ( ) ( ) ax ax + by " d ! x = a 2 + b 2 f x , y ( ) ds L " We have: u = a 2 + b 2 a 2 + b 2 f x , y ( ) ds L ! = 1 a 2 + b 2 f x , y ( ) ds L ! Adding this particular solution to the corresponding solution to the homogenous equation we have: u = 1 a 2 + b 2 f x , y ( ) ds L ! + g bx " ay ( )...
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This note was uploaded on 04/08/2008 for the course MATH 3028 taught by Professor Desilva during the Spring '08 term at Columbia.

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