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1. Solve x:2x1, xj:x,+xz, xli):,x1[):3 72sin21 ec052t ecosZt 2. a. Show that it]: _l e sin2t .Filz): 2 _1 are solutions of E'2AR, A:[_1 _2].

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1. Solve x":2x1, xj:x,+xz, xli):,x1[):3 72"sin21 e"c052t e"cosZt 2. a. Show that it]: _l e sin2t .Filz): 2 _1 are solutions of E'2AR, A:[_1 _2]. b. Find the particular solution if x[)=[] . 3. Solve E':A?c,A: '2 '7 71 4 - _ o 4 4151 ':A ,A: 0V9 X X 4 0] 5. Suppose you are writing a program to model the growth of 2 competing strains of a bacteria that start with 100 cells each. Bacteria 1 normally grows in population by 30% each hour, but bacteria 2 dx secretes a chemical that inhibits it so its growth rate is dg=0.3x]O.Sx1 . Bacteria 2 normally grows in population by 10% per hour. but bacteria 1 secretes a chemical that d boosts its growth rate to i=0.1x'+0.1x2 . Once either bacteria goes to a population oi 0, the cf: other bacteria continues on its own with its original growth rate. Which bacteria goes to 0, if either does. and how long does it take? 3 2 6 B. Solve E'ZAEH'K: 2 1 2 1 2 4 1 9 2 _._ _Cl 2 H [I 7. Solve x 7AX,A7 0 72 I] 0 1 (95} B. Solve '2 1 2 2 71 9.Solve '=A},A=2 1 71 1. *= 2e"-2e-' 2e"+e-' 2. a. Determine v, ' and Av, and show equality. Then do the same for v, ' and Av, . b. x=v,+3v2 3. X= CI -6 + C2 7e -31 4. X=c, Cost + c2 sin 4t [sin 4t] -cos 4t 5. bacteria 2 goes to 0 after - hours. 8. *=cie- |1-2:] 1-2t 9. X=C1 -2 +c20 +czer 2 -4t 10. *=Ce 1+2t +cze'[-1 +e-10t*+312+3t 10t-6t

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