Homogeneous systems Constant Coefficients: Problem 3 1 pt Consider the systems o

Question

Homogeneous systems Constant Coefficients: Problem 3 1 pt Consider the systems of differential equations dx/dt = 0.4x + 0.5y, dy/dt = 1.5x - 0.6y. For this system, the smaller eigenvalue is and the larger eigenvalue is . Use the phase plotter pplane7.m in MATLAB to determine how the solution curves behave. All of the solution curves run away from 0. Unstable node All of the solution curves converge towards 0. Stable node The solution curves converge to different points. The solution curves race towards zero and then veer away towards infinity. Saddle The solution to the above differential equation with initial values x 0 = 2, y 0 = 9 is x t = , y t = .

Explanation / Answer

1) Noting A the underlying matrix then :
det(A-XI) = (0.4-X)(-0.6-X)-1.5*0.5 = X^2+0.2X-0.99=(X+1.1)(X-0.9)
So the smallest eigenvalue is L1=-1.1 and the bigger is L2=0.9

2) I dont have matlab so i'll let you do this one

3) We need to find the eigenvectors.

Solving AX=-1.1X leads to the system :

0.4a + 0.5b = -1.1a => b = -3a
1.5a -0.6b = -1.1b => b = -3a
So we can take as first eigenvector v1=(1,-3)

Solving AX=0.9X leads to the system :

0.4a + 0.5b = 0.9a => a=b
1.5a-0.6b = 0.9b => a=b
So we can take as second eigenvector v2=(1,1)

So the solution has the form X = c1 e^(-1.1t) ( 1,-3) + c2 e^(0.9t) (1,1)

That is :
x(t) = c1e^(-1.1t) + c2e^(0.9t)
y(t) = -3c1e^(-1.1t) + c2e^(0.9t)

Using initial conditions now :
x(0) = 2 = c1 + c2
y(0) = 9 = -3c1 + c2

Which leads (by subtraction ) to 4c1 = -7 so :
c1 = -7/4 =-1.75 and c2 = 2+7/4 = 15/4 = 3.75

Finally :
x(t) = -1.75e^(-1.1t) + 3.75e^(0.9t)
y(t) = 5.25e^(-1.1t) + 3.75e^(0.9t)

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