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Unformatted text preview: Second Order Differential Equations; Complex Roots of the Characteristic Equation Again considering the massspring system of the previous lec ture, we note that if there is no damping in the system, so that = 0, then we have the differential equation m d 2 y dt 2 + k y = 0 . Here the characteristic equation is m r 2 + k = 0; the roots satisfy r 2 = k m . This equation has no real solutions; the solutions are the complex numbers r 1 = v u u t k m i , r 2 = v u u t k m i , where i has the property i 2 = 1. What form can we expect the solutions to take in this case? Since the term k y is independent of t , we can proceed by the change of variable method introduced earlier for general autonomous second order equations; we set dy dt = z and then have, with K 2 = k m , z dz dy = k m y  K 2 y, or z dz = K 2 y dy. Integrating, we have z 2 2 = K 2 y 2 2 + c. 1 Let us consider the case wherein c > 0 and rename c K 2 c 2 2 . Then we have z = K q c 2 y 2 and the equation dy dt = z becomes dy dt = K q c 2 y 2 , or 1 K c 2 y 2 dy = dt. Now let y = c w and we have 1 K 1 1 w 2 dw = dt 1 K sin 1 w = t + d so that y = c w = c sin( Kt + K d ) and, renaming K d as d , y = c sin( Kt + d ) = c (sin Kt cos d + cos Kt sin d ) . Then letting c 1 = c cos d, c 2 = c sin d , we finally have y ( t, c 1 , c 2 ) = c 1 sin Kt + c 2 cos Kt ; K = v u u t k m ....
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 Spring '10
 ROBINSON
 Equations

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