200_pdfsam_math 54 differential equation solutions odd

200_pdfsam_math 54 differential equation solutions odd - y...

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Chapter 4 35. We look for a particular solution of the form y p ( t )= t s ( A 1 t + A 0 ) e t ,andchoose s = 1 because the auxiliary equation, r 3 + r 2 2=( r 1)( r 2 +2 r +2)= 0 has r =1a sas imp leroo t . Hence, y p ( t )= t ( A 1 t + A 0 ) e t =( A 1 t 2 + A 0 t ) e t y 0 p ( t )= ± A 1 t 2 +(2 A 1 + A 0 ) t + A 0 ² e t y 0 p ( t )= ± A 1 t 2 +(4 A 1 + A 0 ) t +(2 A 1 +2 A 0 ) ² e t y 0 p ( t )= ± A 1 t 2 +(6 A 1 + A 0 ) t +(6 A 1 +3 A 0 ) ² e t y 0 + y 0 2 y =[10 A 1 t +(8 A 1 +5 A 0 )] e t = te t . Equating the corresponding coefficients, we Fnd that 10 A 1 =1 , 8 A 1 +5 A 0 =0 A 1 =1 / 10 , A 0 = 8 A 1 / 5= 4 / 25 y p ( t )= ³ 1 10 t 2 4 25 t ´ e t . EXERCISES 4.5: The Superposition Principle and Undetermined Coefficients Revisited, page 192 1. Let g 1 ( t ):=s in t and g 2 ( t ):= e 2 t .Then y 1 ( t )=cos t is a solution to y 0 y 0 + y = g 1 ( t ) and y 2 ( t )= e 2 t / 3 is a solution to y 0 y 0 + y = g 2 ( t ) . (a) The right-hand side of the given equation is 5 sin t
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Unformatted text preview: y ( t ) = 5 y 1 ( t ) = 5 cos t is a solution to y y + y = 5 sin t . (b) We can express sin t 3 e 2 t = g 1 ( t ) 3 g 2 ( t ). So, by the superposition principle the desired solution is y ( t ) = y 1 ( t ) 3 y 2 ( t ) = cos t e 2 t . (c) Since 4 sin t + 18 e 2 t = 4 g 1 ( t ) + 18 g 2 ( t ), the function y ( t ) = 4 y 1 ( t ) + 18 y 2 ( t ) = 4 cos t + 6 e 2 t is a solution to the given equation. 196...
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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 University of California, Berkeley.

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