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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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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