Question 5.3.1: Find the general solution to the system of differential equa...

The differential equation is given in matrix form by

y′ = Ay = \begin{bmatrix} -1&0 \\3&2  \end{bmatrix} y

After solving the characteristic equation det(A − λI) = 0, we know that the eigenvalues of A are \lambda _{1} = −1       and        \lambda _{2} = 2 with corresponding eigenvectors

v_{1} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}         and          v_{2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

Hence, the matrix P which diagonalizes A (see Theorem 2 of Sec. 5.2) is

P= \begin{bmatrix} 1 &0\\ -1&1 \end{bmatrix}        with      P^{-1}=  \begin{bmatrix} 1 &0\\ 1&1 \end{bmatrix}

The related uncoupled system is then given by

w′= P^{-1} APw

= \begin{bmatrix} 1 &0\\ 1&1 \end{bmatrix} \begin{bmatrix} -1 &0\\ 3&2 \end{bmatrix} \begin{bmatrix} 1 &0\\ -1&1 \end{bmatrix} w

= \begin{bmatrix} -1 &0\\ 0&2 \end{bmatrix} w

whose general solution is

w(t) = \begin{bmatrix} e^{-t} &0\\ 0&e^{2t} \end{bmatrix} w(0)

Hence, the solution to the original system is given by

y(t) = \begin{bmatrix} 1 &0\\ -1&1 \end{bmatrix} \begin{bmatrix} e^{-t} &0\\ 0&e^{2t} \end{bmatrix}\begin{bmatrix} 1 &0\\ 1&1 \end{bmatrix}y(0)

= \begin{bmatrix} e^{-t} &0\\ -e^{-t} + e^{2t}&e^{2t} \end{bmatrix} y(0)

The general solution can also be written in the form

y_{1}(t) = y_{1}(0)e^{-t}       and          y_{2}(t) = −y_{1}(0)e^{-t}  + [y_{1}(0) + y_{2}(0)] e^{2t}

The phase portrait is shown in Fig. 2. The signs of the eigenvalues and the direction of the corresponding eigenvectors help to provide qualitative information about the trajectories in the phase portrait. In particular, notice in Fig. 2 that along the line spanned by the eigenvector v_{1} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} the flow is directed toward the origin. This is so because the sign of \lambda _{1} = −1 is negative. On the other hand, flow along the line spanned by  is v_{2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} away from the origin, since in this case \lambda _{2} = 2 is positive.