Problem 1.
The functions $x(t)$ and $y(t)$ describe the concentration of two substances, respectively.
The first component transforms into the second one with the rate coefficient $k\gt 0$, and the second component decomposes at the rate $\mu\gt 0$. Describe the system.
We have to investigate the system
\begin{align*}
x'(t) &=-k x(t),\\
y'(t) &=k x(t)-\mu y(t);\\
x(0) &=x_0 ,\,y(0)=y_0.
\end{align*}
From the first equation
\[
x(t)=x_0e^{-kt}.
\]
Substituting it into the second equation we get
\begin{align*}
y'(t)+\mu y(t) &=kx_0e^{-kt},\\
y(0) &=y_0.
\end{align*}
Then \[ y(t)=(kx_0 t+C)e^{-kt}. \] Since $y(0)=y_0$ it follows \[ y(t)=(kx_0 t+y_0)e^{-kt}. \] We note that if $x_0\gt 0$ then \[ \frac{x(t)}{y(t)}=\frac{x_0}{kx_0 t+y_0}, \] and \[ \lim_{t\to\infty}\frac{x(t)}{y(t)}=0. \] Case (b). $\boldsymbol{\mu\neq k}$.
Then \[ y(t)=\left(-\frac{kx_0}{k-\mu}e^{-t(k-\mu)} +C \right)e^{-\mu t}. \] Since $y(0)=y_0$ it follows \[ y(t)=\left(-\frac{kx_0 e^{-t(k-\mu)}}{k-\mu} +y_0+\frac{x_0 k}{k-\mu} \right)e^{-\mu t}. \] We note that if $x_0\gt 0$ then \begin{align*} \frac{x(t)}{y(t)} &=\frac{x_0 e^{-t(k-\mu)}}{-\frac{kx_0 e^{-t(k-\mu)}}{k-\mu} +y_0+\frac{x_0 k}{k-\mu}}\\ &=\frac{x_0}{-\frac{k x_0}{k-\mu}+e^{t(k-\mu)}\left(y_0+\frac{k x_0}{k-\mu} \right)}. \end{align*} If $k\gt \mu$ then \[ \lim_{t\to\infty}\frac{x(t)}{y(t)}=0. \] If $k\lt\mu$ then \[ \lim_{t\to\infty}\frac{x(t)}{y(t)}=\frac{\mu-k}{k}.\quad\blacksquare \]
Using the general formula for solution
we obtain
\[
y(t)=e^{-\mu t}\left[kx_0\int e^{\mu t-kt}dt+C \right].
\]
Case (a). $\boldsymbol{\mu=k}$.
Then \[ y(t)=(kx_0 t+C)e^{-kt}. \] Since $y(0)=y_0$ it follows \[ y(t)=(kx_0 t+y_0)e^{-kt}. \] We note that if $x_0\gt 0$ then \[ \frac{x(t)}{y(t)}=\frac{x_0}{kx_0 t+y_0}, \] and \[ \lim_{t\to\infty}\frac{x(t)}{y(t)}=0. \] Case (b). $\boldsymbol{\mu\neq k}$.
Then \[ y(t)=\left(-\frac{kx_0}{k-\mu}e^{-t(k-\mu)} +C \right)e^{-\mu t}. \] Since $y(0)=y_0$ it follows \[ y(t)=\left(-\frac{kx_0 e^{-t(k-\mu)}}{k-\mu} +y_0+\frac{x_0 k}{k-\mu} \right)e^{-\mu t}. \] We note that if $x_0\gt 0$ then \begin{align*} \frac{x(t)}{y(t)} &=\frac{x_0 e^{-t(k-\mu)}}{-\frac{kx_0 e^{-t(k-\mu)}}{k-\mu} +y_0+\frac{x_0 k}{k-\mu}}\\ &=\frac{x_0}{-\frac{k x_0}{k-\mu}+e^{t(k-\mu)}\left(y_0+\frac{k x_0}{k-\mu} \right)}. \end{align*} If $k\gt \mu$ then \[ \lim_{t\to\infty}\frac{x(t)}{y(t)}=0. \] If $k\lt\mu$ then \[ \lim_{t\to\infty}\frac{x(t)}{y(t)}=\frac{\mu-k}{k}.\quad\blacksquare \]
