Question

solve the IVP using the annihilator approach

y⁽³⁾ + 9y' = e^xcos(3x)

y(0) = 2

y'(0) = 1

y''(0) = 1

Answer 1

Given equation is. y" + gy! =e* cos(3x) let D= d dt then y = D3 y and - DY ou D3y+9DY - SO, equation 18. ex Cosc3x) → (D3 + 9D) y er cos(34) The auxiliary equation is 23+ ga=0 aca? +9) 2 = 0 >? = -9 a=0 a = #3i = 0 T 1 = Yc at C₂ Cos(x) + C, Singa) particular solution. Yp e2x cos(34) D3 +9D - e2ҡ Cos (32) (D+2)3 + 9(0+2) - - een Cos(34) D3 +6D² +212 +26 cs sinced wit? + 9 Annihilates Cos (34) CamScanner

so DP = -9. Yp e 22 Cos(32) 11 121 - 28 3D +7 een 4 Cos (3x) 9 DP-49 e 2x (3D + 7) cos(3x) DE-9 520 2x 520 e² [-9 sin(3x) + 7 Cos(3x)] The general solution 1. Yc typ 7e2n Cos(32) y = ( + C₂ Cospu) + C₂ sinx) + 520 ezre sin (3x) 520 y! - -3 sin (34) (₂ + 3(3 Cos(32) + 7 + Free ezze S20 (2605(3) - 3 sin (3x)) - 9224 são e zu (2 sin (34) + 3 cos (3x) 2n 520 e ? -962 Cos (31) - 963 sin (34) + 7 tulee [ 4605(38) - 65inc3a) – 6C0513) – 9 sin(3x)] cea [ [ 4 sin(3x) +3 60513«) + 6653x) g sina (3x)] - 9 e 2x 520 CS Scanned with CamScanner

using initial conditions. yo) - C +2 + 7 520 - 2 y'(o) 3C3 + = 27 - .14 520 = 1 S20 y" (0) -962 14 11 BI S20 520 we hare. 1033 cit C₂ = 520 C3 533 1560 (2 .615 4680 디 1033 615 4680 - S20 1033 615 4680 9912 3304 520 - 4680 156 Hence the solution es. 826 39 41 312 Los (3x) + 533 sin (34) 1560 Cos (34) - +7 e 2x S20 CS Scanned with CamScanner 9 S20 24 sin (34)