Question

1. Solve the following differential equations using the method of variation of pa- rameters (for first order equations), or any other acceptable method: cos x + (sin x) y = 1 Ans: y = sin c +C cosx. (b) (x+1) 7 + (x + 2) y = 2xe-* Ans: (x+1)e’y = 22 + C. 2 + (3x + 1) y = e-32 Ans: y = e-3x + Cx-1e-3r.

Please use ONLY the variation of parameters method if possible. Thank you!

Answer 1

+(sinx)y COS x dy XP +(tamxly secx (N) dx het + (tanx)y o CH) dx dy tank dx Integrating bothside we ge hag191bogcosx! + hag K d=K Casx Let -2) g=KX) cosx K C) SinK -C3 dx From equation CI), (2) (3) k(x) CDSK K(x) Sinx +ftanx). Ke) CDSx =secx K CUSthx + Kcrsinx= secx k'(x) CoSx KCX)=Secx sec'x dx kix) KOx)= tanx tc solution is ginen by Theme for the gix Kex). COSX (C+ tamx ). Cosx. yix=C cosx sinx

x1)dy + (Z+2) 님 = xp 2xe . + +9) 2 xe (xt1) CN) xP x+1) dy + dx t1) ut CH) (2t9) y dP integrating boths ide we g4 xt1) In 191- n Ixt! + C. In 1y1= -In |Xt1|-X +C Kex x+1) hat K(x) e (2) ke (2t1) K()e 3) (xtU From equation () 2) P (3) k'CX) (x1) 2x e (x11) KOx) + (t2) K()ek t)2 (xt1) xt12 KOe (X1) (X+1) KCz)e 2xe (211) KITJEX KOrIe 2x e Cx+1D 1L

K'Cy=2x K (3) Thare forx is ginen by slution t1) ce + xtl) or (x-t1) 2+c y xt) e e 3x dy (3xt1 = dy(3xt) y (N dx (H) (x4 het dx - (3+1)dr integ rating bothside se get dy 3 x кe У 3 3x KZ) KO) C3) - 3 Kory e3x 1 dn from equation (, 2)3) 3x e K'ae 3 Kce K(Je(3+) Kx)e

-3X e k'cx)e3x 11 dx There fore the solution is ginen by yeI-(XtC)e e ce3 = (2)R 3x TT