197_pdfsam_math 54 differential equation solutions odd

# 197_pdfsam_math 54 differential equation solutions odd - A...

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Exercises 4.4 Substituting y , y 0 ,and y 0 into the original equation, after some algebra we get [ 26 A 1 t +(8 A 1 13 A 0 )] e 3 t = 2 te 3 t 26 A 1 = 2 , 8 A 1 13 A 0 =0 A 1 =1 / 13 , A 0 =8 / 169 . Therefore, y p ( t )= ± t 13 + 8 169 ² te 3 t . 21. The nonhomogeneous term of the original equation is te 2 t . Therefore, a particular solution has the form x p ( t )= t s ( A 1 t + A 0 ) e 2 t . The corresponding homogeneous diFerential equation has the auxiliary equation r 2 4 r +4=( r 2) 2 =0 . S ince r = 2 is its double root, s is chosen to be 2. Thus a particular solution to the nonhomogeneous equation has the form x p ( t )= t 2 ( A 1 t + A 0 ) e 2 t = ( A 1 t 3 + A 0 t 2 ) e 2 t . We compute x 0 p = ( 3 A 1 t 2 +2 A 0 t ) e 2 t +2 ( A 1 t 3 + A 0 t 2 ) e 2 t , x 0 p =(6 A 1 t +2 A 0 ) e 2 t +4 ( 3 A 1 t 2 +2 A 0 t ) e 2 t +4 ( A 1 t 3 + A 0 t 2 ) e 2 t . Substituting these expressions into the original diFerential equation yields x 0 p 4 x 0 p +4 x p =( 6 A 1 t +2 A 0 ) e 2 t +4 ( 3 A 1 t 2 +2 A 0 t ) e 2 t +4 ( A 1 t 3 + A 0 t 2 ) e 2 t 4 ( 3 A 1 t 2 +2 A 0 t ) e 2 t 8 ( A 1 t 3 + A 0 t 2 ) e 2 t +4 ( A 1 t 3 + A 0 t 2 ) e 2 t =( 6 A 1 t +2 A 0 ) e 2 t = te 2 t
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Unformatted text preview: A = 0 and A 1 = 1 / 6. Therefore x p ( t ) = t 3 e 2 t / 6 is a particular solution to the given nonhomogeneous equation. 23. The right-hand side of this equation suggests that y p ( θ ) = θ s ( A 2 θ 2 + A 1 θ + A ). We choose s = 1 because r = 0 is a simple root of the auxiliary equation, r 2 − 7 r = 0. Therefore, y p ( θ ) = θ ( A 2 θ 2 + A 1 θ + A ) = A 2 θ 3 + A 1 θ 2 + A θ ⇒ y p ( θ ) = 3 A 2 θ 2 + 2 A 1 θ + A ⇒ y p ( θ ) = 6 A 2 θ + 2 A 1 . 193...
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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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