Question

Question 2 Find the general solution to 21 2 -2 X1 X2 1 4 32 Enter Answer D Question 3 Find the solution to the initial value problem X1 1 -2 21 2] [3 (2:0) = [] X1(0) 22(0) X2 3 6 X2 Enter Answer

Question 2 Find the general solution to 21 2 -2 X1 X2 1 4 32 Enter Answer D Question 3 Find the solution to the initial value problem X1 1 -2 21 2] [3 (2:0) = [] X1(0) 22(0) X2 3 6 X2 Enter Answer

Answer 1

-2 4):6:10 2-d -2 1 2 -(1-1) 12 Salution (2) Given equation is Here eigen values of the coefficient matrix is given by 20 = (2-1)(4-1) +2 -0 urd 22-69 +10 2 = 6+)36-40 67 36-40 - 3+1 To find the eigenvector for the first eigenvalue 2.-3+i; we need to solve the following system 1-i Using the second equation we get, n+(1-1)^2 = 0 So, the first eigenvector is, ñ -(1-2)22 -Iti (for 12 = 1) N₂ The solution corresponding to this eigenvalue and eigen vector is alti (3+1) (3+i)t X(t) = - - 2 1)()-(3) 커 2) = (-1*')

est (cost risint) =p3t/-cost-sint Juice ( tiest/Cost-sint sint Cost - U(Z) + i v(t) The general solution to the given system is then 3t/-Cost-sint sint +6e3t/Cost-sint 21 (t) 3t cil-cost-sint) +CZ (Cost-sint) e [x2(t) C Cost +G sint
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