# HW6 - MA341003 Homework 6 Solution 4.4 1. No. t-1 et is not...

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MA341–003 Homework 6 Solution 4.4 1. No. t - 1 e t is not a product of polynomials and exponentials. 3. Yes. Since 3 t = e ln 3 t = e t ln 3 , then we can apply the method of undetermined coeﬃ- cients to this equation. 5. No. sec θ = 1 / cos θ it not a sine/cosine, a polynomial, or an exponential. 11. The characteristic equation for the corresponding homogeneous equation is 2 r 2 + 1 = 0, with the roots r = ± 1 2 i . These roots are not linked with the inhomogeneity 9 e 2 t , thus we can look for a particular solution in the form Ae 2 t . Hence, z p ( t ) = Ae 2 t , z 0 p ( t ) = 2 Ae 2 t , z 00 p ( t ) = 4 Ae 2 t . Then, we substitute these expressions into the equation, which yields 2(4 Ae 2 t ) + Ae 2 t = 9 e 2 t , 9 Ae 2 t = 9 e 2 t . Therefore, A = 1. 13. The characteristic equation for the corresponding homogeneous equation is r 2 - r +9 = 0, with the roots 1 2 ± 35 2 i . These roots are not linked with the inhomogeneity 3 sin 3 t , thus we can look for a particular solution in the form A cos 3 t + B sin 3 t . Hence, y p ( t ) = A cos 3 t + B sin 3 t, y 0 p ( t ) = - 3 A cos 3 t + 3 B sin 3 t, y 00 p ( t ) = - 9 A cos 3 t - 9 B sin 3 t. Substituting these expressions into the inhomogeneous equation, we have: - 9 A cos 3 t - 9 B sin 3 t + 3 A sin 3 t - 3 B cos 3 t + 9 A cos 3 t + 9 B sin 3 t = 3 sin 3 t, 3 B cos 3 t + 3 A sin 3 t = 3 sin 3 t, A = 1 , B = 0 . Consequently,

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## This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.

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HW6 - MA341003 Homework 6 Solution 4.4 1. No. t-1 et is not...

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