251_pdfsam_math 54 differential equation solutions odd

251_pdfsam_math 54 differential equation solutions odd - t...

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Review Problems 5. The roots of the auxiliary equation, 6 r 2 11 r +3=0,are r 1 =3 / 2and r 2 =1 / 3. Thus, y ( t )= c 1 e r 1 t + c 2 e r 2 t = c 1 e 3 t/ 2 + c 2 e t/ 3 is a general solution. 7. Solving the auxiliary equation, 36 r 2 +24 r + 5 = 0, we Fnd r = 24 ± p 24 2 4(36)(5) 2(36) = 1 3 ± 1 6 i. Thus a general solution is given by y ( t )= ± c 1 cos ² t 6 ³ + c 2 sin ² t 6 ³´ e t/ 3 . 9. The auxiliary equation, 16 r 2 56 r +49 = (4 r 7) 2 = 0, has a double root r =7 / 4. Therefore, e 7 t/ 4 and te 7 t/ 4 are two linearly independent solutions, and a general solution is given by y ( t )= c 1 e 7 t/ 4 + c 2 te 7 t/ 4 =( c 1 + c 2 t ) e 7 t/ 4 . 11. This equation is a Cauchy-Euler equation. Using the approach discussed in Problem 38, Exercises 4.3, we make the substitution
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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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