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Unformatted text preview: Exercises 4.3 To satisfy the initial conditions, we solve the system 2 = y (0) = c 1 17 / 2 = y (0) = ( c 1 / 2) (5 c 2 / 2) ⇒ c 1 = 2 c 2 = 3 . Therefore, the answer is y = e t/ 2 [2 cosh(5 t/ 2) 3 sinh(5 t/ 2)] . EXERCISES 4.3: Auxiliary Equations with Complex Roots 2. The auxiliary equation in this problem is r 2 + 1 = 0, which has roots r = ± i . We see that α = 0 and β = 1. Thus, a general solution to the differential equation is given by y ( t ) = c 1 e (0) t cos t + c 2 e (0) t sin t = c 1 cos t + c 2 sin t. 4. The auxiliary equation, r 2 10 r + 26 = 0, has roots r = 5 ± i . So, α = 5, β = 1, and y ( t ) = c 1 e 5 t cos t + c 2 e 5 t sin t is a general solution. 6. This differential equation has the auxiliary equation r 2 4 r + 7 = 0. The roots of this auxiliary equation are r = ( 4 ± √ 16 28 ) / 2 = 2 ± √ 3 i . We see that α = 2 and β = √ 3. Thus, a general solution to the differential equation is given by w ( t ) = c 1 e 2 t cos √ 3 t + c 2 e 2 t sin √ 3 t . 8. The auxiliary equation for this problem is given by 4 r 2 + 4 r + 6 = 0 ⇒ 2 r 2 + 2 r + 3 = 0 ⇒ r = 2 ± √ 4 24 4 = 1 2 ± √ 5 2 i. Therefore, α = 1 / 2 and β = √ 5 / 2, and a general solution is given by y ( t ) = c 1 e t/ 2 cos √ 5 t 2 ! + c 2 e t/ 2 sin √ 5 t 2 ! . 10. The associated auxiliary equation, r 2 + 4 r + 8 = 0, has two complex roots, r = 2 ± 2 i . Thus the answer is y ( t ) = c 1 e 2 t cos 2 t + c 2 e 2 t sin 2 t....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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