Differential Equations | Part IA, 2001

Consider the linear system

z˙+Az=h\dot{\mathbf{z}}+A \mathbf{z}=\mathbf{h}


z(t)=(x(t)y(t)),A=(1+a211+a),h(t)=(2costcostsint)\mathbf{z}(t)=\left(\begin{array}{c} x(t) \\ y(t) \end{array}\right), \quad A=\left(\begin{array}{cc} 1+a & -2 \\ 1 & -1+a \end{array}\right), \quad \mathbf{h}(t)=\left(\begin{array}{c} 2 \cos t \\ \cos t-\sin t \end{array}\right)

where z(t)\mathbf{z}(t) is real and aa is a real constant, a0a \geq 0.

Find a (complex) eigenvector, e, of AA and its corresponding (complex) eigenvalue, lll \underline{l}. Show that the second eigenvector and corresponding eigenvalue are respectively e\overline{\mathbf{e}} and lˉ\bar{l}, where the bar over the symbols signifies complex conjugation. Hence explain how the general solution to ()(*) can be written as

z(t)=α(t)e+αˉ(t)e,\mathbf{z}(t)=\alpha(t) \mathbf{e}+\bar{\alpha}(t) \overline{\mathbf{e}},

where α(t)\alpha(t) is complex.

Write down a differential equation for α(t)\alpha(t) and hence, for a>0a>0, deduce the solution to ()(*) which satisfies the initial condition z(0)=0\mathbf{z}(0)=\underline{0}.

Is the linear system resonant?

By taking the limit a0a \rightarrow 0 of the solution already found deduce the solution satisfying z(0)=0\mathbf{z}(0)=0 when a=0a=0.

Typos? Please submit corrections to this page on GitHub.