Paper 2, Section II, A

Differential Equations
Part IA, 2010

(a) Find the general solution of the system of differential equations

(x˙y˙z˙)=(121101121)(xyz)\left(\begin{array}{l} \dot{x} \\ \dot{y} \\ \dot{z} \end{array}\right)=\left(\begin{array}{rrr} -1 & 2 & -1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)

(b) Depending on the parameter λR\lambda \in \mathbb{R}, find the general solution of the system of differential equations

(x˙y˙z˙)=(121101121)(xyz)+2(λ1λ)e2t,\left(\begin{array}{l} \dot{x} \\ \dot{y} \\ \dot{z} \end{array}\right)=\left(\begin{array}{rrr} -1 & 2 & -1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)+2\left(\begin{array}{r} -\lambda \\ 1 \\ \lambda \end{array}\right) e^{2 t},

and explain why (2)(2) has a particular solution of the form ce2t\mathbf{c} e^{2 t} with constant vector cR3\mathbf{c} \in \mathbb{R}^{3} for λ=1\lambda=1 but not for λ1\lambda \neq 1.

[Hint: decompose (λ1λ)\left(\begin{array}{c}-\lambda \\ 1 \\ \lambda\end{array}\right) in terms of the eigenbasis of the matrix in (1).]

(c) For λ=1\lambda=-1, find the solution of (2) which goes through the point (0,1,0)(0,1,0) at t=0t=0.