2. Solution initial condition; y1(0)=2, y2(0)=3 Then y1(0)=4C1-C2=2, and y2(0)=C1+C2=3. We obtain C1=1 and C2=2

(a) The matrix of coefficients A and column matrix of constants B are  1 3 1   2 A  2 5 1 and B    5  1 2 3  6 It is found that |A| = –3  0. Thus Cramer’s rule can be applied. We get  2 3 1  1  2 1  1 3  2 A1    5 5 1 A2  2  5 1 A3  2 5  5        6 2 3  1 6 3  1 2 6 giving A1  3, A2  6, A3  9 。Cramer’s rule now gives A1 3 A 6 A 9 x1    1, x2  2   2, x3  3  3 A 3 A 3 A 3

2. Solution initial condition; y1(0)=2, y2(0)=3 Then y1(0)=4C1-C2=2, and y2(0)=C1+C2=3. We obtain C1=1 and C2=2

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