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209_pdfsam_math 54 differential equation solutions odd

209_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.5 37. Clearly, r = 0 is not a root of the auxiliary equation, r 3 2 r 2 r + 2 = 0. (One can find the roots, say, using the factorization r 3 2 r 2 r + 2 = ( r 2)( r 1)( r + 1), but they are not needed for the form of a particular solution: the only important thing is that they are different from zero.) Therefore, a particular solution has the form y p ( t ) = A 2 t 2 + A 1 t + A 0 . Substitution into the original equation yields y p 2 y p y p + 2 y p = (0) 2(2 A 2 ) (2 A 2 t + A 1 ) + 2( A 2 t 2 + A 1 t + A 0 ) = 2 A 2 t 2 + ( A 1 2 A 2 ) t + ( A 0 A 1 4 A 2 ) = 2 t 2 + 4 t 9 . Equating the coefficients, we obtain 2 A 2 = 2 , 2 A 1 2 A 2 = 4 , 2 A 0 A 1 4 A 2 = 9 A 2 = 1 , A 1 = 3 , A 0 = 1 . Therefore, y p ( t ) = t 2 + 3 t 1. 39. The auxiliary equation in this problem is r 3 + r 2 2 = 0. By inspection, we see that
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