Question

IV. Determine the form for yp but do NOT evaluate the constants. 1. y" - 5y' + 6y = ex cos 2x + e2x(3x + 4) sin x (ans. this is #21(a) in sec. 3.6) 2. y" - 3 y' - 4 y = 3 e2x + 2 sin x - 8eXcos 2x (ans. Yp = Ae2x + B cos x + C sin x + De* cos 2x + E e sin 2x) V. Solve by variation of parameters. 1. y" - 4y = 2e2x (ans. y = C1 e - 2x + C2 e 2x + 2 x e2x) 2. y" - y - 2 y = e 3x (ans. y = cie -* + C2 e 2x + % e 3x) 3. Y" + 4y = 3 csc 2x, 0<x</2 (ans. y = Ci cos 2x + C2 sin 2x + 3% (sin 2x) In sin 2x - 3/2 x cos 2x) 4. Y" + 4y' + 4y = x2 e-2x, x>0 (ans. y = C1 e - 2x + C2 x e -2% - e -2* In x)

Answer 1

Solution IU (1) y" - 5y +6y= e cos2x+e (3x+4) Sinx The chare de ristic equation is m2 5m+6 25-24 5² 2 5 ± 1 m- = m = 3, m = 2 3% 2x e e So, the fundamental set of solution is Now, by method of undetermined Coafficient Y a 2x A é cos2x+Bê sinza + é (Cx+D) Sinx - 22 tē "LEX+f) cosa 2x 8 e cos 22 42 2) y" - 3y - 4y = 6 + 2 Sink The charederistic equation is m²-3m-4507m = 4, m =-1 So, the fundamental set of solution is de since, fundamental set of solution is not bart of non-homogeneous term, so by method of undetermined Coefficient 2a Ур - не + BLOSX + (sinxt De cosat e sinaa