1. (a) Let $x_{s}, t_{s}$ denote scales for displacement & time respectively. Define nondimensional variables $\tilde{t}=\frac{t}{t_{s}}, \tilde{x}=\frac{x}{x_{s}}$. Note that $\tilde{x}(\tilde{t})=x(\tilde{t}t_{s})/x_{s}$. Therefore using chain rule $\frac{d}{d \tilde{t}} \tilde{x} (\tilde{t}) = \frac{t_{s}}{x_{s}} \frac{d}{dt} x(t), \; \frac{d^{2}}{d \tilde{t}^{2} }\tilde{x}(\tilde{t}) = \frac{t_{s}^{2}}{x_{s}} \frac{d^{2}}{dt^{2}} x(t)$ The initial value problem becomes$\frac{m}{t_{s}^{2}} \frac{d^{2}}{d \tilde{t}^{2}} \tilde{x} (\tilde{t}) + \frac{c }{t_{s}} \frac{d}{d \tilde{t}} \tilde{x}(\tilde{t}) + k \; \tilde{x}(\tilde{t})=0, \quad \tilde{x}(0) = 0, \frac{d}{d\tilde{t}} \tilde{x}(0) = \frac{t_{s}}{x^{s}}v$Note that $m/t_{s}^{2}=c/t_{s}$ gives $t_{s} = m/c$ so $m/t_{s}^{2}=c/t_{s}=c^{2}/m$. So choose $t_{s} = m/c,\; x_{s}=mv/c$. Dividing through by $c^{2}/m$ gives $\frac{d^{2}}{d \tilde{t}^{2}} \tilde{x} (\tilde{t}) + \frac{d}{d \tilde{t}} \tilde{x}(\tilde{t}) + \tilde{k} \; \tilde{x}(\tilde{t})=0, \quad \tilde{x}(0) = 0, \frac{d}{d\tilde{t}} \tilde{x}(0) = \frac{t_{s}}{x^{s}}v=1,$where $\tilde{k}= km/c^{2}$. (b) The auxiliary equation is reads $\lambda^{2} + \lambda+\tilde{k}=0$ and has solutions $\lambda_{1,2} = -\frac{1}{2} \pm \frac{1}{2} \sqrt{ 1- 4\tilde{k} }$ Depending on the parameters we have the following solution regimes: - Underdamped case: If $1-4\tilde{k} <0$ then we have the linearly independent solutions $\tilde{x_{1}}(\tilde{t})= e^{p\tilde{t}} \cos(\omega \tilde{t})$, $\tilde{x_{2}} (\tilde{t}) = e^{p\tilde{t}} \sin(\omega \tilde{t})$, where $p = -\frac{1}{2}$ and $\omega=\frac{1}{2} \sqrt{ 4 \tilde{k} -1 }$. Hence the general solution is given by $\tilde{x}(\tilde{t})=l_{1} \tilde{x_{1}}(\tilde{t}) + l_{2}\tilde{x_{2}}$, $l_{1},l_{2} \in \mathbb{R}$. Accounting for initial conditions we obtain the specific solution $\tilde{x}(t)= \frac{e^{p\tilde{t}}}{\omega} \sin(\omega \tilde{t}) $ - Over-damped case: If $1-4\tilde{k} >0$ then we have the linearly independent solutions $\tilde{x_{1}}(\tilde{t}) = e^{\lambda_{1}\tilde{t}},\; \tilde{x_{2}} (\tilde{t})=e^{\lambda_{2} \tilde{t}}$. Hence the general solution is given by $\tilde{x}(\tilde{t})=l_{1} \tilde{x_{1}}(\tilde{t}) + l_{2}\tilde{x_{2}}$, $l_{1},l_{2} \in \mathbb{R}$. Accounting for initial conditions we obtain the specific solution $\tilde{x}(\tilde{t})= \frac{e^{\lambda_{1} \tilde{t}} - e^{\lambda_{2}\tilde{t}}}{\lambda_{1}-\lambda_{2}} $ - Critically damped case: If $1-4\tilde{k}=0$ then $\lambda_{1,2}=-\frac{1}{2}$ so the general solution is $\tilde{x}(\tilde{t})= l_{1}e^{-\frac{1}{2}\tilde{t}} +l_{2}\tilde{t} e^{-\frac{1}{2}\tilde{t}},\; l_{1},l_{2} \in \mathbb{R}$. Accounting for initial conditions we obtain the specific solution $\tilde{x}(\tilde{t})=\tilde{t}e^{-\frac{1}{2}\tilde{t}}$. 2. (a) Let $w=v^{-1}$, so $\frac{dw}{dt} = -v^{-2} \frac{dv}{dt}$. The initial value problem becomes $\frac{dw}{dt} - w=- \epsilon, \quad w(0)=1. $Hence $\begin{align} w(t) &= e^{t} \left(1- \epsilon \int_{0}^{t} e^{-\tilde{t}} \, d\tilde{t} \right) \\ \implies w(t) &= e^{t} (1+ \epsilon(e^{-t} -1)) \\ \implies v(t) &= \frac{e^{-t}}{1+\epsilon(e^{-t}-1)} \end{align}$ (b) Assuming a solution of the form $v(t) = v_{0}(t)+\epsilon v_{1}(t).$ Substituting $v(t)$ and its derivative gives $v_{0}'(t) +\epsilon v_{1}'(t) = -(v_{0}(t)+ \epsilon v_{1}(t)) + \epsilon (v_{0}(t)+\epsilon v_{1}(t))^{2} $ with initial conditions $v_{0}(0)=1, v_{1}(0)=0.$Collecting powers of $\epsilon$ gives the series of linear ODEs $v_{0}'(t) = -v(t), \; v_{1}'(t)= -v_{1}(t) +v_{0}(t)^{2}.$Solving them gives $v_{0}(t)=e^{-t}, \; v_{1}(t)=e^{-t}-e^{-2t}.$So our approximate solution is given by $v(t)=e^{-2t} +\epsilon (e^{-t}- e^{-2t}).$ 3. (a) By the law of mass action the corresponding differential equations read $\begin{align} \frac{d}{dt} a(t) &= -2 k_{1} a(t)^{2} b(t)^{3} + 2k_{2} x(t) y(t)^{6} \\ \frac{d}{dt} b(t) &= -3k_{1} a(t)^{2}b(t)^{3 } + 3k_{2} x(t)y(t)^{6} \\ \frac{d}{dt} x(t) &= k_{1} a(t)^{2} b(t)^{3} - k_{2} x(t) y(t)^{6} \\ \frac{d}{dt} y(t) &= 6k_{1}a(t)^{2} b(t)^{3}- 6k_{2}x(t) y(t)^{6} \end{align}$where $a(t), b(t), x(t), y(t)$ denote the time-dependent concentrations of $A,B,X,Y$ respectively. (b)The chemical reaction is given by $CO+NO_{3}^{-} \stackrel{k}{\to} NO_{2}+CO_{2}$. By the law of mass action the corresponding differential equations read $\begin{align} \frac{d}{dt} w(t) &= -kw(t)x(t) \\ \frac{d}{dt} x(t) &= -kw(t)x(t) \\ \frac{d}{dt } y(t) &= kw(t)x(t) \\ \frac{d}{dt} z(t) &= kw(t)x(t) \end{align}$where $w(t), x(t), y(t), z(t)$ denote the time-dependent concentrations of $CO, NO_{3}^{-1}, NO_{2}, CO_{2}$ respectively. (c)The DEs are satisfied by the coupled equations $I+S \stackrel{\beta}{\to} 2I, \quad I \stackrel{\gamma}{\to} R.$