\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We have \begin {align*} \DiffFrac {x - y}{t} & = \DiffFrac {x}{t} - \DiffFrac {y}{t} \\ & = (-x + 3y + u) - (x + y + u) \\ & = -2x + 2y \\ & = -2 (x - y). \end {align*}
This is a differential equation for \(x - y\) in terms of \(t\), and hence it solves to \[ x - y = A e^{-2t}. \]
If \(x = y = 0\) for some \(t > 0\), then it must be the case that \(A = 0\), giving \(x - y = 0\), and \(x = y\).
Therefore, for \(t = 0\), we must also necessarily have \(x_0 = y_0\).
Given that \(x_0 = y_0\), we must have \(x = y\) for all \(t > 0\). Hence, \begin {align*} \DiffFrac {x}{t} & = -x + 3x + u \\ \DiffFrac {x}{t} & = 2x + u \\ \frac {\Diff x}{2x + u} & = \Diff t \\ \ln \abs *{2x + u} & = 2 t + C \\ 2x + u & = A e^{2t}. \end {align*}
Since at \(t = 0\), \(x = x_0\), we must have \(A = 2 x_0 + u\), and hence \[ 2x + u = (2x_0 + u) e^{2t}, \] and rearranging gives \[ u = \frac {2 (x_0 e^{2t} - x)}{1 - e^{2t}}. \]
The particle is at origin at time \(t = T > 0\), and hence \(x = y = 0\) for \(t = T\), and hence \[ u = \frac {2 x_0 e^{2T}}{1 - e^{2T}}. \]
This ensures the particle is at origin as well since this ensures the particle is at \(x = 0\) for \(t = T\), and \(y = x\) so \(y = 0\) as well.
Consider \(\DiffFrac {x}{t} + \DiffFrac {z}{t} - 2 \DiffFrac {y}{t}\), and we have \begin {align*} \DiffFrac {x + z - 2y}{t} & = \DiffFrac {x}{t} + \DiffFrac {z}{t} - 2 \DiffFrac {y}{t} \\ & = (4y - 5z + u) + (x - 2y + u) - 2 (x - 2z + u) \\ & = 4y - 5z + u + x - 2y + u - 2x + 4z - 2u \\ & = -x - z + 2y, \end {align*}
and hence \[ x + z - 2y = A e^{-t}. \]
Since the particle is at the origin at some time \(t > 0\), we must have \(A = 0\), and hence \[ x + z - 2y = 0, \] which means \(y = \frac {x + z}{2}\) for all time \(t\).
At time \(t = 0\), \(y_0 = \frac {x_0 + z_0}{2}\), and so \(y_0\) is the mean of \(x_0\) and \(z_0\).
Since \(2y = x + z\), we must have \[ \DiffFrac {x}{t} = 2 (x + z) - 5z + u = 2x - 3z + u, \] and \[ \DiffFrac {z}{t} = x - (x + z) + u = -z + u. \]
Hence, considering \(\DiffFrac {x}{t} - \DiffFrac {z}{t}\), we have \begin {align*} \DiffFrac {x - z}{t} & = \DiffFrac {x}{t} - \DiffFrac {z}{t} \\ & = (2x - 3z + u) - (-z + u) \\ & = 2 (x - z), \end {align*}
which gives \[ x - z = A e^{2t}. \]
Since the particle is at the origin for some \(t > 0\), we must have \(A = 0\). This means \(x = z\) for all \(t\), and further we have \(x = y = z\) for all \(t\) since \(2y = x + z\).
At \(t = 0\), this means \(x_0 = y_0 = z_0\) as desired.
Given that \(x_0 = y_0 = z_0\), all previous parts still apply, since the boundary condition of \(2y = x + z\) and \(x = z\) holds for \(t = 0\). Hence, \(x = y = z\) for all \(t\), and \begin {align*} \DiffFrac {x}{t} & = -x + u \\ \frac {\Diff x}{x - u} & = - \Diff t \\ \ln \abs *{x - u} & = -t + C \\ x - u & = A e^{-t}. \end {align*}
At \(t = 0\), \(x = x_0\), we must have \(A = x_0 - u\), and hence \[ x - u = (x_0 - u) e^{-t}, \] and rearranging gives \[ u = \frac {x_0 e^{-t} - x}{1 - e^{-t}}. \]
The particle is at origin at a time \(t = T > 0\), and hence \(x = y = z = 0\) for \(t = T\), and hence \[ u = \frac {x_0 e^{-T}}{1 - e^{-T}} = \frac {x_0}{1 + e^{T}}. \]
This ensures the particle is at origin as well since this ensures the particle is at \(x = 0\) for \(t = T\), and \(x = y = z\), so \(y = z = 0\) as well.