371_pdfsam_math 54 differential equation solutions odd

# 371_pdfsam_math 54 differential equation solutions odd - 5...

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Exercises 6.3 D 3 ( D 3)( D 2 5 D +6)[ y ]= D 3 ( D 3) 2 ( D 2)[ y ]=0 . This last equation has the associated auxiliary equation r 3 ( r 3) 2 ( r 2) = 0 , which has roots r = 0, 0, 0, 3, 3, 2. Thus, a general solution to the diferential equation associated with this auxiliary equation is y ( x )= c 1 e 2 x + c 2 e 3 x + c 3 xe 3 x + c 4 x 2 + c 5 x + c 6 . The homogeneous equation, y 0 5 y 0 +6 y = 0, associated with the original problem, is the same as in Problem 21 (with u replaced by y ). ThereFore, the solution to the homogeneous equation associated with the original problem is y h ( x )= c 1 e 2 x + c 2 e 3 x . Since a general solution to this original problem is given by y ( x )= y h ( x )+ y p ( x )= c 1 e 2 x + c 2 e 3 x + y p ( x ) and since y ( x ) must be oF the Form y ( x )= c 1 e 2 x + c 2 e 3 x + c 3 xe 3 x + c 4 x 2 + c 5 x + c 6 , we see that y p ( x )= c 3 xe 3 x + c 4 x 2 + c
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Unformatted text preview: 5 x + c 6 . 25. ±irst, we rewrite the equation in operator Form, that is, ( D 2 − 6 D + 9 ) [ y ] = sin 2 x + x ⇒ ( D − 3) 2 [ y ] = sin 2 x + x . In this problem, the right-hand side is a sum oF two Functions. The ²rst Function, sin 2 x , is annihilated by ( D 2 + 4), and the operator D 2 annihilates the term x . Thus A := D 2 ( D 2 + 4) annihilates the Function sin 2 x + x . Applying this operator to the original equation (in operator Form) yields D 2 ( D 2 + 4)( D − 3) 2 [ y ] = D 2 ( D 2 + 4)[sin 2 x + x ] = 0 . (6.9) 367...
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