Question 3.5.1: Consider the system of differential equations given by x´= A...

By inspection, since A is a triangular matrix, we see that λ = −2 is a repeated eigenvalue of A with multiplicity 2. From this, we deduce that v_{1} = [1   0]^{T} is a corresponding eigenvector, and therefore one solution to x´= Ax is x_{1} = c_{1}e^{−2t }[1   0]^{T}. However, A lacks a second linearly independent eigenvector associated with λ = −2; therefore, we need to find a second real linearly independent solution to the system in order to determine the general solution to x´= Ax. In this example, we are fortunate that the system is only partially coupled and that therefore we may solve the system directly by using techniques for single differential equations from chapter 2.

In particular, noting that the second equation in the system is \acute{x_{2}} =−2x_{2},
it follows immediately that the solution to this single differential equation is x_{2}(t ) = ce^{−2t} . Substituting this result into the equation \acute{x_{1}}= −2x_{1} + x_{2}, it remains for us to solve the single nonhomogeneous linear first-order differential equation

\acute{x_{1}}=−2x_{1} +ce^{−2t}

Applying our understanding of such equations from section 2.3, via the integrating factor v(t ) = e^{2t} we know that

x_{1}(t ) =\frac {1}{e^{2t}} \int {e^{2t} · ce^{−2t} dt}=e^{−2t} (ct +k)

To summarize, with x_{1}(t ) and x_{2}(t ) as the components of x(t ), we have found that a solution to the system is

x(t ) =\begin{bmatrix} x_{1}(t ) \\ x_{2}(t )\end{bmatrix}

=\begin{bmatrix} e^{−2t} (ct +k) \\ ce^{−2t}\end{bmatrix}                   (3.5.1)

If we factor this expression to write x(t ) as a linear combination of two vectors in order to more clearly identify the role of the constants in (3.5.1), we see

x(t ) = k\begin{bmatrix} e^{−2t} \\ 0\end{bmatrix}+c \begin{bmatrix} te^{−2t} \\ e^{−2t}\end{bmatrix}

In this form, two key observations can be made. First, each individual vector in (3.5.2) may be verified to be a solution to the given system. Moreover, these two vectors are linearly independent. Hence, (3.5.2) is the general solution to the given system.