285_pdfsam_math 54 differential equation solutions odd

285_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 5.2 Thus we have 2 2 (AB )[y ] := A B [y ] = j =0 2 2 j aj D j i=0 i bi D i [y ] 2 2 j i=0 := j =0 2 aj D 2 bi D [y ] := i=0 2 j =0 2 aj D bi D i [y ] = j =0 i=0 2 aj D j bi D i [y ] = i=0 j =0 2 2 i j =0 2 2 j bi D i aj D j [y ] 2 i j =0 = i=0 bi D aj D [y ] aj D j j =0 =: i=0 bi D aj D j [y ] =: i=0 bi D i [y ] = B A[y ] =: (BA)[y ]. (b) We have {(A + B ) + C } [y ] := (A + B )[y ] + C [y ] := (A[y ] + B [y ]) + C [y ] = A[y ] + (B [y ] + C [y ]) =: A[y ] + (B + C )[y ] =: {A + (B + C )} [y ] and {(AB )C } [y ] := (AB ) C [y ] := A B C [y ] =: A (BC )[y ] =: {A(BC )} [y ]. (c) Using the linearity of differential operators, we obtain {A(B + C )} [y ] := A (B + C )[y ] := A B [y ] + C [y ] = A B [y ] +A C [y ] =: (AB )[y ] + (AC )[y ] =: {(AB ) + (AC )} [y ]. 41. As it was noticed in Example 2, we can treat a “polynomial” in D , that is, an expression of the form p(D ) = n i=0 ai D i , as a regular polynomial, i.e., p(r ) = n i=0 ai r i , while per- forming arithmetic operations. Hence, the factorization problem for p(D ) is equivalent to the factorization problem for p(r ), which is the same as finding its roots. (a) r = −3 ± 32 − 4(−4) −3 ± 5 = = −4, 1 2 2 ⇒ D 2 + 3D − 4 = (D + 4)(D − 1). 281 ...
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