Question

(17 points) Match the third order linear equations with their fundamental solution sets. 1. y" + 3y" + 3y + y=0 2. Y" - 8y" + 15y = 0 3. y" - y" - y' + y = 0 4. ty" - y = 0 5. y" + y = 0 6. y" – 3y" + y - 3y = 0 est A. 1, e3t, B. e3t, cos(t), sin(t) c.1, t, 13 D. 1, cos(t), sin(t) E. e , te, te F.et, tet, e-

Answer 1

So the solution is y = net A et + Bettetet tet "Яз- The wronokian ent et tet -et et tet tet Y=e-t, Y2 = et tet W7,142143) ent et et tetter to so y . Ye. Y3 are linearly independent Hence the equation has furecimentad Det F. 11 4. ler ty" -yla o y". =P tp' -Po 5 becomes, 1 de dt ť P=.0 ling Integrating factor - é logt t multiplying & (4) both sides of cib by O ta Integrating, f y" ДА [but p= 4 + B { integrating Y'E AL? 2

1 = at Bl + C (again integrating y = HE The Wronnkian, 42=ti 13= 13 t til at 12:43) 372 - 6t o a -6t to i, Yz 43 are lirenaly Hence the equation has fondamental net C. Lirenody in deterdenta

5. y!!! O 2 AD at -> (o? ologi The Qumiliony equation in :> m(m?+1): 0 :>m=0,+ i met me 0 So the solution is 4 nt a cost of c sint let y Yo = cost, f 3= sint The wronskian 1144.79.13) cost Sint 0 -Sint Cost -cost -Sint Sint - so @quation 1.70 y, Y2 Y are linearly independent Hence the has fundamental net . D. 6. y" - 34" + yl-3=0 -> (- 30? + 0-3 D-394-0, Da ग auril equation .bm2 of m. 2 m? (m. 3) 41(ma 3)O (m? +1) ( m3): 0 The ma

So the solution in y = Aest to cost te sint let y = 0 3t 92= cost, Yo - sint The wronohian 'W (174, 42, 43) e3b COS2 Sint 303L Sinf Cost q3t COS Sint so t Y ,G2, Yz are independent Hence the equation has fundamental B. linearly set

UI y Wheren 24 +37" :+2 3ylty (13..+302 +38-+)yed euriliary equation in m3 +3m² tomt 1 = 0 dt The (m+1)3 > ma 47,-- The solution solution of in y = nært at Blert + Ble-t tethet Let yere 92=tet Y za tret This Wronnkian et tett t? et Let et let ztethet W(YY₂93)= e -tattetet untet 428 € tthetzet so 2. ť o yu. 92, 93 are line.cody indo pendent. Hence the equation has fundamental solution ARTE " By " +154'=0 -(D3 - 80? +150 +15D) y O, Where da auxilio liary in 8m² +15m=8 -> m (m2. 8m +15)= 0 2 24 equation m3

im (m-5) (m-3). c. mo 0,3,5 the solution is "50 y - At Best test Y za est Leo % = 1, Y = 0 3t The ronrakian W/4, 4₂, 43): 0.34 O Best 55 25e5t qe3t 30e8t so are re linearly Y., 92, 93 indefendent. Hence the equation has fundamenter net A. 2. y" y" -yhty=0 Dit l.)y=0 equation in motz m? (mod) (m-1) = 0 => [m? +) (-4)= a (03 -D? Da 2- ar The auxiliary m3 m2 -->> 2 > lilil