193_pdfsam_math 54 differential equation solutions odd

193_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.4 EXERCISES 4.4: Nonhomogeneous Equations: The Method of Undetermined Coefficients, page 186 1. We cannot use the method of undetermined coefficients to Fnd a particular solution because of the t 1 term, which is not a polynomial. 3. Rewriting the right-hand side in the form 3 t = e (ln 3) t = e rt ,where r = ln 3, we conclude that the method of undetermined coefficients can be applied. 5. Since sec θ =1 / cos θ , we cannot use the method of undetermined coefficients. 7. Given equation is not an equation with constant coefficients. Thus the method of undeter- mined coefficients cannot be applied. 9. The roots of the auxiliary equation, r 2 +3=0,are r = ± 3 i . Since they are di±erent from zero, we look for a particular solution of the form y p ( t ) A . Substitution into the original equation yields ( A ) 0 +3 A = 9 3 A = 9 A = 3 . Thus, y p ( t ) ≡− 3 is a particular solution to the given nonhomogeneous equation.
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Unformatted text preview: 11. The auxiliary equation in this problem, 2 r 2 + 1 = 0, has complex roots. Therefore, e 2 t is not a solution to the corresponding homogeneous equation, and a particular solution to the original nonhomogeneous equation has the form z p ( t ) = Ae 2 t . Substituting this expression into the equation, we Fnd the constant A . 2 ( Ae 2 t ) + Ae 2 t = 2 ( 4 Ae 2 t ) + Ae 2 t = 9 Ae 2 t = 9 e 2 t ⇒ A = 1 . Hence, z p ( t ) = e 2 t . 12. This equation is a linear Frst order di±erential equation with constant coefficients. The corresponding homogeneous equation, 2 x + x = 0, can be solved by the methods of Chapter 2. Alternatively, one can use the result of Problem 21 in Section 4.2. Either approach yields 189...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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