264_pdfsam_math 54 differential equation solutions odd

# 264_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 From the second equation we get x 0 y 0 =0 ( x y ) 0 x ( t ) y ( t )= c 1 x ( t c 1 + c 2 e 2 t , and a general solution is given by x ( t c 1 + c 2 e 2 t ,y ( t c 2 e 2 t . 5. Writing this system in operator notation yields the system ( D 1)[ x ]+ D [ y ]=5 , D [ x ]+( D +1)[ y ]=1 . (5.1) We will frst eliminate the ±unction x ( t ), although we could proceed just as easily by eliminat- ing the ±unction y ( t ). Thus, we apply the operator D to the frst equation and the operator ( D 1) to the second equation to obtain D ( D 1)[ x D 2 [ y ]= D [5] = 0 , ( D 1) D [ x ] ( D 1)( D y ( D 1)[1] = 1 . Adding these two equations yields { D ( D 1) ( D 1) D } [ x ± D 2 ( D 2 1) ² [ y 0 · x +1 · y =1 y ( t )=1 . To fnd the ±unction x ( t ), we will eliminate y ±rom the system given in (5.1). There±ore, we
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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