181_pdfsam_math 54 differential equation solutions odd - Exercises 4.3 49(a To nd the roots of the auxiliary equation p(r:= 3r 3 18r 2 13r 19 = 0 one

# 181_pdfsam_math 54 differential equation solutions odd -...

• Notes
• JudgeComputerQuail10099
• 1

This preview shows page 1 out of 1 page.

Exercises 4.3 49. (a) To find the roots of the auxiliary equation, p ( r ) := 3 r 3 + 18 r 2 + 13 r 19 = 0, one can use Newton’s method or intermediate value theorem. We note that p ( 5) = 9 < 0 , p ( 4) = 25 > 0 , p ( 2) = 3 > 0 , p ( 1) = 17 < 0 , p (0) = 19 < 0 , p (1) = 15 > 0 . Therefore, the roots of p ( r ) belong to the intervals [ 5 , 4], [ 2 , 1], and [0 , 1], and we can take r = 5, r = 2, and r = 0 as initial quesses. Approximation yields r 1 ≈ − 4 . 832, r 2 ≈ − 1 . 869, and r 3 0 . 701. So, a general solution is given by y ( t ) = c 1 e r 1 t + c 2 e r 2 t + c 3 e r 3 t = c 1 e 4 . 832 t + c 2 e 1 . 869 t + c 3 e 0 . 701 t . (b) The auxiliary equation, r 4 5 r 2 + 5 = 0, is of quadratic type. The substitution s = r 2 yields s 2 5 s + 5 = 0 s = 5 ± 5 2 r = ± s = ± 5 ± 5 2 . Therefore, r 1 = 5 5 2 1 . 176 , r 2 = 5 + 5 2 1 . 902 , r 3 = r 1 , and r 4 = r 2 are the roots of the auxiliary equation, and a general solution to y (ıv) 5 y + 5 y = 0 is given by y ( t ) = c 1 e r 1 t + c 2 e r 1 t + c 3 e r 2 t + c 4 e r 2 t . (c) We can use numerical tools to find the roots of the auxiliary fifth degree polynomial equation r 5 3 r 4 5 r 3 + 15 r 2 + 4 r 12 = 0. Alternatively, one can involve the rational
• • • 