Question

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Problem 4. Solve the linear non-homogeneous equation for r = r(t) with initial conditions ë+ 4.+ 5x = 10t + 13, 2(0) = 2, i(0) = 0.

Answer 1

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Paget x +44 +50= lot + 13 (0)=2 x(0)=0 +4dx+5x = at lot +13 +2 (²+40+5)2 = lot + 13 For CF. 2²+40+5= 0 m2 +4m +5=0 ax² + br +c=0 al 6=4c=5 on +bx+c=0 x=-b= 15-uac za m= – b = 162-yal za = -4 ± √ (4)² - 4x1x5 2 x since when m=&tiß CF=ast a Cs Bat Gempt = -4 = 16-20 2 = -4 I FY = -4 I 2 2 = -2t e 2 2 CF- 2 (c. cast + Casink] -1 et PI = I lot + 13 D²+4075 + 131 = 10 0² + UDTS +40+5 = 10 it & 13 - lo SL1+ 40 +2²) ot 4xoto

Page 2 ant rest of team higher/ order Deivative - Ithathin this 2) = 2 (1+ (y + 2) + + 13 +2 L.- t o test one on By using Binomics = 21*-**- 4x0 + 0)+ (1+x)? = 2 + + + -- = 2t - 8 + 12 / 2 = at + g = att C.S=3(t) = cft PI XG) = 32+ [ cicopt +68m+]+2t+1 0 - warty - 2+ /-C Bont + C Cont) + Ce cost +6287) (- 2e2t]+2 Since xco) = 2 x(0)=0 using aco)=2 in 0 - t=0 x = 2 0 2 = È LC Cosot Casino] + 2x0+) 2= [colt /6 =3 => c=1] - using & (o)= 0 in 2 d =0 x=0 @=> m(o)= e [- ci sinot (26@so) + Caso +CSmo] (-28) +2 Now

Page 3 - cat c, 4+2 x 1+2 o = C2 - 20, +2 o = C2 - 2x1+2 = ca - 272 Gal stence using 14=1 K2=0) in ea a(t) = eet Lacost & Ca Sint]+27 +1 xt) = ezt a cost tox sint] +att) at) = ezt copt + at +1) 18 Required so Please like it