Question

Consider the initial value problem: 2' - 2+ 2(0) = (*) a. Form the complementary solution to the homogeneous equation. -e (t) = 21 +02 b. Construct a particular solution by assuming the form zp(t) = ae+ bt+c and solving for the undetermined constant vectors a, b, and c. 2p(t) = c. Solve the original initial value problem. 31(t) ) - 22(0)

Answer 1

داو -1 t Jout :) -( ezt (0340) t's all tt -> (1 let lip UCO). Du, +12=t -0) Hy'= -x, teet Ut D&2 get (1) XD -(11) we we get (024) 02-1) u Then aux m2-1=0 =) mati Get te ēt, where as q are constant = - D(t) - €2+ e2t auxilary equation 1 so die be particular solution. (1-eat) suppose from=emu Then The nups Fol 1 mu lip D2-1 - e²t (1) Fom emu 02-1 D2-1 et if fimto 3

solution is Therefore the complete Mia tulip Get te et nie eht -|- 3 From (2 we (1) we get Ng = t - Du -> U2 = ēt 3 Now (11) =t-getthet + 2/3 et -get tq + t + 2 ezt M (0) = 0 => +4-1 - $ -- => G+ 2 & Kp0= 17 -9 +4 +} G-₂ (111) tawi we we get 29 = 1 » 4= ez - + + 3 = 5 / 5 Therefore al, = 2 et + q et eht - ett et + feet + t +ş eat. Ś -(iv) 2 From Tung -|- & N a) H.(t)= a 5/6ēt 516 et + 1½ et q ) ét My(t) = a, td et et where &, = (3/6) & d₂ = ( de

-! - 2 3 り Mp(t) : E + 30 21 (1 3 + t t १ ) (6) 23. Where az up f() = a ²t + bt te ) 2/3/ ier (1) १६.) - (2) + = (%) - (i) ()- () + + + ९) (4) AD