5. Consider two general solutions of two different linear systems of equations X'(t) = AX(t): A. X(t)=c₁e B. X(t)=c₁e -2 -4 -t [2] The phase portraits for these solutions are below: -2 [2] 0 2 -4t -4t + c₂e XX 4 -2 + c₂e -t ] 0 2 4 Which phase portrait corresponds to general solution A? To B? How do you know?

Write the system given system of differential equation in matrix form. Y₁' = 3y₁ + y2 + 3 sin(4x) Y2 = y12y2 + 4 sin(4x) [] Y2 = Y1 S + sin(4x)

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This is the first part of a two-part problem. Let P=[-: 1 5₁(t) = [(41) 5₂(t) = - sin(4t). a. Show that y₁ (t) is a solution to the system ÿ' = Pÿ by evaluating derivatives and the matrix product y(t) = = 0 [1] -4 Enter your answers in terms of the variable t. -4 sin(4t) -4 cos(4t)] ÿ₁ (t) [181-18] b. Show that y₂ (t) is a solution to the system ÿ' = Pÿ by evaluating derivatives and the matrix product Enter your answers in terms of the variable t. 04] 32(t) = [-28]|2(t) 181-181