377_pdfsam_math 54 differential equation solutions odd

377_pdfsam_math 54 differential equation solutions odd - a...

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Exercises 6.3 which means that r = 0 is not a root of this polynomial. Thus, for the auxiliary equation (6.16), r = 0 is a root of exact multiplicity m + 2, and so a general solution to (6.15) is given by y ( x ) = c 0 + c 1 x + · · · + c m +1 x m +1 + Y ( x ) , (6.17) where Y ( x ), being associated with roots of a n r n 1 + a n 1 r n 2 + · · · + a 1 = 0, is a general solution to ( a n D n 1 + a n 1 D n 2 + · · · + a 1 ) [ y ] = 0. (One can write down Y ( x ) explicitly but there is no need in doing this.) On the other hand, the auxiliary equation for the homogeneous equation, associated with (6.14), is r ( a n r n 1 + a n 1 r n 2 + · · · + a 1 ) = 0, and r = 0 is its simple root. Hence, a general solution y h ( x ) to the homogeneous equation is given by y h ( x ) = c 0 + Y ( x ) , (6.18) where Y ( x ) is the same as in (6.17). Since y ( x ) = y h ( x ) + y p ( x ), it follows from (6.17) and (6.18) that y p ( x ) = c 1 x + · · · + c m +1 x m +1 = x ( c 1 + · · · + c m +1 x m ) , as stated. 37. Writing equation (4) in operator form yields
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Unformatted text preview: ( a n D n + a n − 1 D n − 1 + ··· + a ) [ y ] = f ( x ) . (6.19) The characteristic equation, corresponding to the associated homogeneous equation, is a n r n + a n − 1 r n − 1 + ··· + a = 0 . (6.20) Suppose that r = βi is a root of (6.20) of multiplicity s ≥ 0. ( s = 0 means that r = βi is not a root.) Then (6.20) can be factored as a n r n + a n − 1 r n − 1 + ··· + a = ( r 2 + β 2 ) s ( a n r n − 2 s + ··· + a /β 2 s ) = 0 and so a general solution to the homogeneous equation is given by y h ( x ) = ( c 1 cos βx + c 2 sin βx ) + x ( c 3 cos βx + c 4 sin βx ) + ··· + x s − 1 ( c 2 s − 1 cos βx + c 2 s sin βx ) + Y ( x ) , (6.21) 373...
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