Question

Given the Homogeneous Linear Differential Equations with Constant Coefficients, determine the general solution

y(v) + 4y(iv) + 5y“” – 6y' – 4y = 0 y(x) = cte* + c2e-2x + c4e +04e-* + e-* (c4Cos x + C5 Sen x) Answer:

Answer 1

yre)+ tyli) tylio) tsyll_by'-ay (1) = 0 then let y=ema be a solution of (1) from (1), we get ma omema nema +5 m3 ema ma ms.ee ent amt.e =0 → m 5+4 m4 +5 m3 - 6m-4=0.-> (2) Now, we solve @) ... 2 20 m5+ 4 m 45 m² 6m-4 =0 a may m-1) + 5(m-1) + 10m (m-1) +10 m (m-1) + 4 (m-1) = 0 (m-1) (74 +5m3 + 10m +10m+1) =0 (m=1> [m?(m+1) + m2(6+1) + Gm(m+1)+46 (m-1)(m+I) (m3 + 4 m+ +6m + 1 → (m-1) (m+1) 1+2) + 2m (m+ 2) + 2(m+2)) 0 → (m-1) (m+1) (m+2) (m'+ 2m +2) =0 (myxom +3 0. -2 + 74-4.1.2 ma-191,-2, 2 iei, m=-1, 1, -2 -lti, ali >

tience the solution of (1) be y (2) = qe2 + 6 e 2x tg éx tex (ca cosa cosa + cs sima) Where a, 2, 3, 4, GG are arbitrasy constants

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