376_pdfsam_math 54 differential equation solutions odd

376_pdfsam_math 54 differential equation solutions odd - x...

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Chapter 6 is a general solution to the given nonhomogeneous equation. Next, we fnd constants c 1 , c 2 , and c 3 such that the initial conditions are satisfed. DiFerentiation yields y 0 ( x )= 2 c 1 e 2 x +[(2 c 2 + c 3 )cos x +(2 c 3 c 2 )sin x ] e 2 x +(2 x 2 x 2 ) e 2 x 2 x, y 0 ( x )=4 c 1 e 2 x +[(3 c 2 +4 c 3 )cos x +(3 c 3 4 c 2 )sin x ] e 2 x +(2 8 x +4 x 2 ) e 2 x 2 . There±ore, 3= y (0) = c 1 + c 2 +3 , 0= y 0 (0) = 2 c 1 +2 c 2 + c 3 , 0= y 0 (0) = 4 c 1 +3 c 2 +4 c 3 c 1 + c 2 =0 , 2 c 1 +2 c 2 + c 3 =0 , 4 c 1 +3 c 2 +4 c 3 =0 . The solution o± this homogeneous linear system is c 1 = c 2 = c 3 = 0. Hence, the answer is y ( x )= x 2 e 2 x x 2 +3. 35. a 0 =0 ,thenequat ion(4)becomes a n y ( n ) + a n 1 y ( n 1) + ··· + a 1 y 0 = f ( x ) or, in operator ±orm, ( a n D n + a n 1 D n 1 + ··· + a 1 D ) [ y ]= f ( x ) D ( a n D n 1 + a n 1 D n 2 + ··· + a 1 ) [ y ]= f ( x ) . (6.14) Since the operator D m +1 annihilates any polynomial f ( x )=
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Unformatted text preview: x m + ··· + b , applying D m +1 to both sides in (6.14) yields D m +1 D ( a n D n − 1 + a n − 1 D n − 2 + ··· + a 1 ) [ y ] = D m +1 [ f ( x )] = 0 ⇒ D m +2 ( a n D n − 1 + a n − 1 D n − 2 + ··· + a 1 ) [ y ] = 0 . (6.15) The auxiliary equation, corresponding to this homogeneous equation is, r m +2 ( a n r n − 1 + a n − 1 r n − 2 + ··· + a 1 ) = 0 . (6.16) Since a 1 6 = 0, ( a n r n − 1 + a n − 1 r n − 2 + ··· + a 1 )± ± r =0 = a 1 6 = 0 , 372...
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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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