209_pdfsam_math 54 differential equation solutions odd

# 209_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.5 37. Clearly, r = 0 is not a root of the auxiliary equation, r 3 2 r 2 r + 2 = 0. (One can Fnd the roots, say, using the factorization r 3 2 r 2 r +2=( r 2)( r 1)( r + 1), but they are not needed for the form of a particular solution: the only important thing is that they are di±erent from zero.) Therefore, a particular solution has the form y p ( t )= A 2 t 2 + A 1 t + A 0 . Substitution into the original equation yields y 0 p 2 y 0 p y 0 p +2 y p =( 0 ) 2(2 A 2 ) (2 A 2 t + A 1 )+2( A 2 t 2 + A 1 t + A 0 ) =2 A 2 t 2 +( A 1 2 A 2 ) t +( A 0 A 1 4 A 2 )=2 t 2 +4 t 9 . Equating the coeﬃcients, we obtain 2 A 2 =2 , 2 A 1 2 A 2 =4 , 2 A 0 A 1 4 A 2 = 9 A 2 =1 , A 1 =3 , A 0 = 1 . Therefore, y p ( t )= t 2 +3 t 1. 39. The auxiliary equation in this problem is r 3 + r 2 2 = 0. By inspection, we see that r =0is not a root. Next, we Fnd that
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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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