194_pdfsam_math 54 differential equation solutions odd

# 194_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 x h ( t )= Ce t/ 2 . So, the homogeneous equation does not have a polynomial solution (other than x ( t ) 0), and we look for a particular solution to the nonhomogeneous equation of the form x p ( t )= A 2 t 2 + A 1 t + A 0 . Substitution into the original diFerential equation yields 2 x 0 p ( t )+ x p ( t )=2(2 A 2 t + A 1 )+ A 2 t 2 + A 1 t + A 0 = A 2 t 2 +(4 A 2 + A 1 ) t +(2 A 1 + A 0 )=3 t 2 . By equating coeﬃcients we obtain A 2 =3 , 4 A 2 + A 1 =0 A 1 = 12 , 2 A 1 + A 0 =0 A 0 =24 . Therefore, a particular solution is x p ( t )=3 t 2 12 t + 24. 13. The right-hand side of the original nonhomogeneous equation suggest us the form y p ( t )= t s ( A cos 3 t + B sin 3 t ) for a particular solution. Since the roots of the auxiliary equation, r 2 r + 9 = 0, are diFerent from 3 i , neither cos 3 t nor sin 3 t is a solution to the corresponding homogeneous equation.
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Unformatted text preview: Therefore, we can choose s = 0, and so y p ( t ) = A cos 3 t + B sin 3 t, y p ( t ) = − 3 A sin 3 t + 3 B cos 3 t, y p ( t ) = − 9 A cos 3 t − 9 B sin 3 t. Substituting these expressions into the original equation and equating the corresponding co-eﬃcients, we conclude that ( − 9 A cos 3 t − 9 B sin 3 t ) − ( − 3 A sin 3 t + 3 B cos 3 t ) + 9 ( A cos 3 t + B sin 3 t ) = 3 sin 3 t ⇒ − 3 B cos 3 t + 3 A sin 3 t = 3 sin 3 t ⇒ A = 1 , B = 0 . Hence, the answer is y p ( t ) = cos 3 t . 190...
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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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