251_pdfsam_math 54 differential equation solutions odd

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t = e s and obtain dx ds = dx dt dt ds = t dx dt , d 2 x ds 2 = d ds ² dx ds ³ = dx ds + t 2 d 2 x dt 2 , ⇒ t 2 d 2 x dt 2 + 5 x = ² d 2 x ds 2 − dx ds ³ + 5 x = d 2 x ds 2 − dx ds + 5 x = 0 . The axiliary equation to this constant coeﬃcient 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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online