374_pdfsam_math 54 differential equation solutions odd

374_pdfsam_math 54 - 2 x 3 c 2 e 3 x − 3 c 3 e − 3 x 2 x y x = 4 c 1 e − 2 x 9 c 2 e 3 x 9 c 3 e − 3 x 2 Therefore − 2 = y(0 = c 1 c 2 c

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Chapter 6 Since the operator D 3 annihilates the nonhomogeneous term in the original equation and r = 0 is not a root of the auxiliary equation, we seek for a particular solution of the form y p ( x )= C 0 x 2 + C 1 x + C 2 . Substituting y p into the given equation (for convenience, in operator form) yileds ( D 3 +2 D 2 9 D 18 C 0 x 2 + C 1 x + C 2 ² = 18 x 2 18 x +22 0+2(2 C 0 ) 9[2 C 0 x + C 1 ] 18 ± C 0 x 2 + C 1 x + C 2 ² = 18 x 2 18 x +22 ⇒− 18 C 0 x 2 +( 18 C 1 18 C 0 ) x +( 18 C 2 9 C 1 +4 C 0 )= 18 x 2 18 x +22 . Equating coefficients, we obtain the system 18 C 0 = 18 , 18 C 1 18 C 0 = 18 , 18 C 2 9 C 1 +4 C 0 =22 C 0 =1 , C 1 =0 , C 2 = 1 . Thus, y p ( x )= x 2 1and y ( x )= y h ( x )+ y p ( x )= c 1 e 2 x + c 2 e 3 x + c 3 e 3 x + x 2 1 is a general solution to the original nonhomogeneous equation. Next, we satisfy the initial conditions. DiFerentiation yields y 0 ( x )= 2 c 1 e 2
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Unformatted text preview: 2 x + 3 c 2 e 3 x − 3 c 3 e − 3 x + 2 x, y ( x ) = 4 c 1 e − 2 x + 9 c 2 e 3 x + 9 c 3 e − 3 x + 2 . Therefore, − 2 = y (0) = c 1 + c 2 + c 3 − 1 , − 8 = y (0) = − 2 c 1 + 3 c 2 − 3 c 3 , − 12 = y (0) = 4 c 1 + 9 c 2 + 9 c 3 + 2 ⇒ c 1 + c 2 + c 3 = − 1 , − 2 c 1 + 3 c 2 − 3 c 3 = − 8 , 4 c 1 + 9 c 2 + 9 c 3 = − 14 . Solving this system, we ±nd that c 1 = 1, c 2 = − 2, and c 3 = 0, and so y ( x ) = e − 2 x − 2 e 3 x + x 2 − 1 gives the solution to the given initial value problem. 370...
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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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