Question

We consider the non-homogeneous problem y" + 2y + 2y = 40 sin(2x) First we consider the homogeneous problem y" + 2y + 2y = 0: 1) the auxiliary equation is ar? + br +C = 242r42 = 0. 2) The roots of the auxiliary equation are 141-14 Center answers as a comma separated list). 3) A fundamental set of solutions is -1 .-1xco) Center answers as a comma separated list. Using these we obtain the the complementary solution y + for arbitrary constants and Next we seek a particule solution y, of the non-homogeneous problem y" + 2y + 2y 40 sin(2x) using the method of undetermined coefficients (See the link below for a help sheet 4) Apply the method of undetermined coefficients to find y, y2 and a particular solution: y + y. Finally you are asked to use the We then find the general solution as a sum of the complementary solution y = y + general solution to solve an IVP. 5 Given the initial conditions (0) = -9 and 7(0) = -8 find the unique solution to the IVP

Answer 1

The given equation is y"+2y'+ 2 y = 40 Sin (2x) .......(1). The reduced (or homogeneous) equation of (ig is y"+ 2y'+ 2y = 0 -----(2), Let y= emx be the trial solution of (2). Then auxiliary equation of (2) is m²+2m +2= of m= - 2 I 4 - 4.2.1 2 = -1 m=alti # i . The complementary function (or a fundamental set of solutions) of ③ Tye = ex {G. cosk & C. Sin x}, Where a, G are arbitrary constants To find the particular integral of a yo use undetermined coefficient. Since shiso (right hand side) of a contains sinax so we take yo an to = A.cos2x + B. sin are ; where A & B are constants to be determined Yo = R(-A Sin 2x + B cus2a) yő =-4 ( A CNS 2X + B Sin 2x).

putting the value of yo, yo & Yp in a we get - 4 A cus 2x - 4 B sin an - 4 Asinan + 4B cos 2 a 4 2 A cos are + 2B Sinaa = 40 sinar HIERRAS MODE cos 2x 1- 4 A+ 4B + 2A)+ sin 2x (-4B - 4 A + 2B) = 40 Sinza * (-2A+ 4B) evs 2 x + (-4A - 2B). Sin2x = 40 sinza Compairing the coefficients of sin af cos a n we get – 2A + 4 B=0 and – 4A-RB = 40. > A = 2B and -2A - B = 20. A = 2B and – 4B-B = 20. A=2B and B = – 4. > A = – 8 and BE - 4. y = - 8 cos2x – 4 Sinax .. The general solution of (1) is y = Ye + y = ē (G. cosa + C. Sina) - 8 cos 2x - 4 sin ax - y = - e* (G. cosx + G. Sinx) + ex (Glinx " +C cosa) + 16 Sinax - 8 cosaa. Given y(0)=-9 → 9-8=-9 → 9 = -1 Abo, y'(o)= - 8 - G+C₂ - 8 = - 8 → G =C2 = -1. " The unique solution to the lvp is ly= ēx (- eosx-sina) - 8 cosm - 4 Sinan