187_pdfsam_math 54 differential equation solutions odd

187_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.3 (c) The fourth order auxiliary equation r 4 + 13 r 2 + 36 = 0 can be reduced to a quadratic equation by making a substitution s = r 2 . This yields s 2 + 13 r + 36 = 0 s = 13 ± 169 144 2 = 13 ± 5 2 . Thus, s = ( 13+5) / 2 = 4 or s = ( 13 5) / 2 = 9, and the solutions to the auxiliary equation are r = ± 4 = ± 2 i and r = ± 9 = ± 3 i . A general solution, therefore, has the form y ( t ) = c 1 cos 2 t + c 2 sin 2 t + c 3 cos 3 t + c 4 sin 3 t . 31. (a) Comparing the equation y + 16 y = 0 with the mass-spring model (16) in Example 4, we conclude that the damping coefficient b = 0 and the stiffness constant k = 16 > 0. Thus, solutions should have an oscillatory behavior. Indeed, the auxiliary equation, r 2 + 16 = 0, has roots r = ± 4 i , and a general solution is given by y ( t ) = c 1 cos 4 t + c 2 sin 4 t . Evaluating y ( t ) and substituting the initial conditions, we get
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Unformatted text preview: y (0) = c 1 = 2 , y (0) = 4 c 2 = 0 ⇒ c 1 = 2 , c 2 = 0 ⇒ y ( t ) = 2 cos 4 t . (b) Positive damping b = 100 and stiFness k = 1 imply that the displacement y ( t ) tends to zero, as t → ∞ . To con±rm this prediction, we solve the given initial value problem explicitly. The roots of the associated equation are r = − 100 ± √ 100 2 − 4 2 = − 50 ± √ 2499 . Thus the roots r 1 = − 50 − √ 2499 and r 2 = − 50 + √ 2499 are both negative. A general solution is given by y ( t ) = c 1 e r 1 t + c 2 e r 2 t ⇒ y ( t ) = c 1 r 1 e r 1 t + c 2 r 2 e r 2 t . 183...
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