\(% 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}}\)
Since \(\dot {y} = -2(y - z)\), differentiating both sides with respect to \(t\) gives \begin {align*} \ddot {y} & = -2 \dot {y} + 2 \dot {z} \\ & = -2 \dot {y} + 2 (- \dot {y} - 3 z) \\ & = -4 \dot {y} - 6z \\ & = -4 \dot {y} - 3(\dot {y} + 2y) \\ & = -7 \dot {y} - 6y, \end {align*}
and hence \[ \ddot {y} + 7\dot {y} + 6y = 0. \]
The auxiliary equation \[ \lambda ^2 + 7 \lambda + 6 = 0 \] gives roots \[ \lambda _{1} = -1, \lambda _{2} = -6, \] and hence \[ y = Ae^{-t} + Be^{-6t}. \]
Hence, \[ \dot {y} = -Ae^{-t} - 6Be^{-6t}, \] and therefore, \begin {align*} z & = \frac {\dot {y} + 2y}{2} \\ & = \frac {(-Ae^{-t} - 6Be^{-6t}) + 2(Ae^{-t} + Be^{-6t})}{2} \\ & = \frac {Ae^{-t} - 4Be^{-6t}}{2} \\ & = \frac {1}{2}A e^{-t} - 2B e^{-6t}. \end {align*}
This set of general solution \[ (y, z) = \left (Ae^{-t} + Be^{-6t}, \frac {1}{2}A e^{-t} - 2B e^{-6t}\right ), \] is exactly what is desired.
\(y(0) = 5\) and \(z(0) = 0\) gives the system of linear equations \[ \left \{ \begin {aligned} A + B & = 5, \\ \frac {1}{2}A - 2B & = 0. \end {aligned} \right . \]
This solves to \((A, B) = (4, 1)\). Hence, \[ z_1(t) = 2e^{-t} - 2e^{-6t}. \]
\(z(0) = z(1) = c\) gives the system of linear equations \[ \left \{ \begin {aligned} \frac {1}{2}A - 2B & = c, \\ \frac {1}{2e}A - \frac {2}{e^6}B & = c, \end {aligned} \right . \implies \left \{ \begin {aligned} A - 4B & = 2c, \\ e^5 A - 4B & = 2e^6c. \end {aligned} \right . \]
Hence, \[ A = \frac {2c(e^6 - 1)}{e^5 - 1}, \] and therefore \begin {align*} B & = \frac {A - 2c}{4} \\ & = \frac {\frac {2c(e^6 - 1)}{e^5 - 1} - 2c}{4} \\ & = \frac {c}{2} \cdot \frac {(e^6 - 1) - (e^5 - 1)}{e^5 - 1} \\ & = \frac {c e^5 (e - 1)}{2(e^5 - 1)}. \end {align*}
This gives \[ z_2(t) = \frac {c(e^6 - 1)}{e^5 - 1} e^{-t} - \frac {ce^5 (e - 1)}{e^5 - 1} e^{-6t}. \]
Notice that \begin {align*} & \phantom {=} \sum _{n = -\infty }^{0} z_1(t - n) \\ & = \sum _{n = -\infty }^{0} [2e^{-t + n} - 2e^{-6t + 6n}] \\ & = 2\sum _{n = 0}^{\infty } [e^{-t-n} - e^{-6t-6n}] \\ & = 2 \left [e^{-t} \sum _{n = 0}^{\infty } e^{-n} - e^{-6t} \sum _{n = 0}^{\infty } e^{-6n}\right ] \\ & = 2 \left [\frac {e^{-t}}{1 - e^{-1}} - \frac {e^{-6t}}{1 - e^{-6}}\right ] \\ & = \frac {2e}{e - 1} e^{-t} - \frac {2e^6}{e^6 - 1} e^{-6t}. \end {align*}
Hence, \(c\) must be such that \[ \left \{ \begin {aligned} \frac {c(e^6 - 1)}{e^5 - 1} & = \frac {2e}{e - 1}, \\ \frac {2e^6}{e^6 - 1} & = \frac {ce^5 (e - 1)}{e^5 - 1}. \end {aligned} \right . \]
Both solves to precisely \[ c = \frac {2e(e^5 - 1)}{(e - 1)(e^6 - 1)}, \] and hence \[ z_2(t) = \sum _{n = -\infty }^{0} z_1(t - n) \] for this value of \(c\).