372_pdfsam_math 54 differential equation solutions odd

372_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 This homogeneous equation has associated characteristic equation r 2 ( r 2 +4)( r 3) 2 =0 with roots ± 2 i , and double roots r =0and r = 3. Therefore, a general solution to (6.9) is given by y ( x )= c 1 e 3 x + c 2 xe 3 x + c 3 + c 4 x + c 5 cos 2 x + c 6 sin 2 x. (6.10) Since the homogeneous equation, ( D 3) 2 [ y ] = 0, which corresponds to the original equation, has a general solution y h ( x )= c 1 e 3 x + c 2 xe 3 x , the “tail” in (6.10) gives the form of a particular solution to the given equation. 27. Since y 0 +2 y 0 +2 y = ( D 2 +2 D +2 ) [ y ]= ± ( D +1) 2 +1 ² [ y ] , the auxiliary equation in this problm is ( r +1) 2 +1 = 0, whose roots are r = 1 ± i . Therefore, a general solution to the homogeneous equation, corresponding to the original equation, is y h ( x )=( c 1 cos x + c 2 sin x ) e x . Applying the operator D 3 { ( D +1) 2 +1 } to the given equation, which annihilates its right-hand
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Unformatted text preview: side, yields D 3 ± ( D + 1) 2 + 1 ²± ( D + 1) 2 + 1 ² [ y ] = D 3 ± ( D + 1) 2 + 1 ²³ e − x cos x + x 2 ´ = 0 ⇒ D 3 ³ ( D + 1) 2 + 1 ´ 2 [ y ] = 0 . (6.11) The corresponding auxiliary equation, r 3 [( r + 1) 2 + 1] 2 = 0 has a root r = 0 of multiplicity three and double roots r = − 1 ± i . Therefore, a general solution to (6.11) is given by y ( x ) = ( c 1 cos x + c 2 sin x ) e − x + ( c 3 cos x + c 4 sin x ) xe − x + c 5 x 2 + c 6 x + c 7 . Since y ( x ) = y h ( x ) + y p ( x ), we conclude that y p ( x ) = ( c 3 cos x + c 4 sin x ) xe − x + c 5 x 2 + c 6 x + c 7 . 368...
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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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