Question

Plea
2. a) (7 pnts) Solve the second order homogeneous linear differential equation y" - y = 0. b) (6 pnts) Without any solving, explain how would you change the above differential equation so that the general solution to the homogeneous equation will become G cos x + cz sinx. c) (7 pnts) Solve the second order linear differential equation y" - y = 3e2x by using Variation of Parameters.
se solve them in clear hand write . Thankyyyoouu

Answer 1

-> (2+1) { 15+1(4-1) 1% - 3(4-1) +18] =0 = (2+1) (4-x) (5+1-3) = 0 (2 + x)² (4-1) = 0 1=-2, -2, 4 Eigen values of A -2,-2.4. => ane by Let u be и. the eigen vector for = -2. (A-AI) U > 3 -3 3 -3 3 U2 0 - U, -ugt Uz = 0 Uz u, - Ug Uz = 0 U = AI Let V be the eigen vector such that (A-I) v = u => 3V, -3 V2 + 3 V₃ = 1 V = 0, y = 0, VF/ 3 3 -3 -3 3 6 3 6 Va moo Let be the cigen vector for 1 = 4. (A-1I) WA 0 -3 -3 3 3 Wg - Wi-Wat Wg = 0 -6 Wq-Wz=0 3 Wg = Wz W, = 0 O-- Eigen vectors ane

m 2. a) - y = 0 Let y=emu (to) be the troial solution. Auxiliary equation m² - 1 - -> 1 => m = 1 1 The solution 7 ciek & each where ca are arbitrary constants. by The differential equation can be written y"+y=0 c) J"-y = 3 e 24 C.F. = ciek+ , ex P.I 3 e2u 2 D 2u 1 an e² 3 2² - 1 the solution is LR cien + C₂ en te The netone 7 -3 - 3. at A = 3 3 3 -4 6 -6 تی characteristie TAXI1=0 equation 1-1 -3 3 - 5-2 - 6 3 4-1 - 2-2 O ata -5-1 3 3 6 - 6 4-1 → (2+1) 3 11 -5-1 -6 3 4-11 6