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Unformatted text preview: t = e s and obtain dx ds = dx dt dt ds = t dx dt , d 2 x ds 2 = d ds dx ds = dx ds + t 2 d 2 x dt 2 , t 2 d 2 x dt 2 + 5 x = d 2 x ds 2 dx ds + 5 x = d 2 x ds 2 dx ds + 5 x = 0 . The axiliary equation to this constant coecient linear equation is r 2 r + 5 = 0, which has roots r = 1 p 1 2 4(1)(5) 2 = 1 19 2 . Thus, y ( s ) = e s/ 2 c 1 cos 19 s 2 + c 2 sin 19 s 2 247...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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