252_pdfsam_math 54 differential equation solutions odd

# 252_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 is a general solution as a function of s . The back substitution, s =ln t , yields y ( t )= t 1 / 2 c 1 cos 19 2 ln t + c 2 sin 19 2 ln t  . 13. The roots of the auxiliary equation, r 2 + 16 = 0, are r = ± 4 i . Thus a general solution to the corresponding homogeneous equation is given by y h ( t )= c 1 cos 4 t + c 2 sin 4 t. The method of undetermined coeﬃcients suggests the form y p ( t )=( A 1 t + A 0 ) e t for a particular solution to the original equation. We compute y 0 p ( t )=( A 1 t + A 0 + A 1 ) e t ,y 0 p ( t )=( A 1 t + A 0 +2 A 1 ) e t and substitute y 0 p ( t )and y p ( t ) into the given equation. This yields y 0 p +16 y p = ± ( A 1 t + A 0 +2 A 1 ) e t ² +16 ± ( A 1 t + A 0 ) e t ² = te t (17 A 1 t +17 A 0 +2 A 1 ) e t = te t A 1 = 1 17 ,A 0 = 2 289 . Therefore, y p ( t )= ³ t 17 2 289 ´ e t y ( t )= y h ( t )+ y p ( t )= c 1 cos 4 t + c 2 sin 4 t + ³ t 17 2 289 ´ e t . 15. This is a third order homogeneous linear diFerential equation with constant coeﬃcients. Its
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Unformatted text preview: This is a third order homogeneous linear diFerential equation with constant coecients. Its auxiliary equation is 3 r 3 + 10 r 2 + 9 r + 2 = 0. actoring yields 3 r 3 + 10 r 2 + 9 r + 2 = (3 r 3 + 3 r 2 ) + (7 r 2 + 7 r ) + (2 r + 2) = (3 r 2 + 7 r + 2)( r + 1) . Thus the roots of the auxiliary equation are r = 1 and r = 7 p 7 2 4(3)(2) 6 = 2 , 1 3 , and a general solution is given by y ( t ) = c 1 e 2 t + c 2 e t + c 3 e t/ 3 . 248...
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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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