Problem 2.
The functions $u(t)$ and $v(t)$ describe the concentration of two substances, respectively.
The components transform into each other with the rate coefficient $k\gt 0$, and the following system describes the process
\begin{align*}
u'(t)&=ku(t)v(t),\\
v'(t)&=-ku(t)v(t).
\end{align*}
For the sake of simplicity let $k:=1$, and $u(0),v(0)=1$. Approximate the solutions by 3rd order polynomials.
From the system, it follows
\[
u'(0)=1,\quad v'(0)=-1.
\]
Since
\begin{align*}
u''(t)&=ku'(t)v(t)+ku(t)v'(t),\\
v''(t)&=-ku'(t)v(t)-ku(t)v'(t)
\end{align*}
we obtain
\[
u''(0)=0,\quad v''(0)=0.
\]
Similarly,
\begin{align*}
u'''(t)&=ku''(t)v(t)+2ku'(t)v'(t)+ku(t)v''(t),\\
v'''(t)&=-ku''(t)v(t)-2ku'(t)v'(t)-ku(t)v''(t),
\end{align*}
and
\[
u'''(0)=-2,\quad v'''(0)=2.
\]
Hence
\begin{align*}
u(t)&=1+t-\frac{1}{3}t^3+\ldots, \\
v(t)&=1-t+\frac{1}{3}t^3+\ldots. \quad\blacksquare
\end{align*}
