Question

Transform the second-order initial-value proben x"+ 4x + 13x = 40 cost, for [0,1], with X(o)=3, x'(o)=4 into a system of first order initial value problems, and use the Euler nethod with 4=0.2 to approxiuste the solution.

Answer 1

2"+ 4x' +13% = 40 cos(t) Let x = u and x'=v =) * = " and x" = v' =v = 40 colt) – 132 -4x' = 40 cos (t) - 13u - 419 • U - V - flt, u, v) 201 = 40 cosít) -13U- 4 19 = glt, 1, 1) And as x(0)=3 U (0) = 3 and x'(0) = 4 =) 7(0) = 4 i. We have a system of two first order equation W = v 201 - 40 cos (t) – 13 U-40 with u(0)= 3, 8(0)=4 Euler's Method is given as; cscanned with CamScanner

nu Uinti = Unth fltn, Un, Un) Unti = Vn th gltn, Uni Un) tnxe = tnth Here to =0, lo=3, Vo=4 ,h=0.2 given Ist iteration tz = to th= 0+ 0.d = 0.2 U = Uoth f ( to, Ho, Vo) = Hotho = 3 + (0.9) x 4 = 3.8 V = Vo thg 1to, lo, va) = Voth [ 40 609 [to) – 4 Vo -13 Uo) = 4+ (0.2) [ 40 colo) - 47 4 1385] = 4 + 10.2) [ 40x1 – 16 – 39] = 1 and iteration : ty = tith= 0.8+0.2= 0.4 cSScanned with CamScanner

Ug = U,+ h flts, U, V.) =ų thu, = 7.8 + (0-2) x1 = 4 V = Vithg Ito, Us, Vi) = wi th [ 40 cos (ti) - 40,- 13 4] = 1+ (0.2) [4ocos (0.2) – 4X1 – 13% 3.8] = -1.839467377 zod iteration : tz= tgth=0.4+0.2=0.6 Urz= Mg + hf (tz, Ug, Vs) = 4+ (0.9) (-1.839467377) = 3.632106525 Vz = Vq thg (tai Ug, V) = V + h [ 40 c3( tia) – 13Xuy - 4-v] =:-3.399405523 CS$canned with CamScanner

4th iteration : +4= tz th = 0.6+0.2= 0.8 UA = Uz + h fltz, U3, 03) = Uz th Uz = 2.952225420 VA = V₂ thg (t3, 43, 43) 5 -3.399405523 + 5.6)-13X3.632106525 - 4x(-3.3994 E -3.044953385 5th iteration : t5 = t4th = 0.8+0.2= 1 Ug = Wq + h fltp, UA, VA) = Uqth Uq = (2.952225420) + (0-2) (-3.0 4495 3385) = 2.343234 743 : 2(2) U(1) * U5 = .34 3234743 1 x (t) 2.343234743 Scanned with CamScanner