Problem 1.
A tank contains $10$ L of water in which initially $y_0$ kg of salt is dissolved. Brine runs in $2$ L per minutes containing $30$% of salt per liter and runs out $2$ L per minutes. The mixture in the tank is kept uniform by stirring.
(a) Determine the amount of salt $y(t)$ in the tank at all times $t>0$.
Show that $\displaystyle{\lim_{t\to\infty}}y(t)=y_{*}$ independently of $y_0$ where $y_{*}$ is the equilibrium solution of the problem.
(b) If initially there is no salt in the tank, i.e., $y(0)=y_0=0$, determine $y(5)$.
(a) Since
\[
y'(t)=\text{salt inflow rate} -\text{salt outflow rate},
\]
we get
\[
y'(t)=0.6-0.2 y(t).
\]
The equilibrium solution is
\[
y_{*}=3. \]
From the differential equation
\[
\frac{1}{0.6-0.2 y}\,dy=1\,dt.
\]
Integrating both sides we obtain
From the initial condition
\[
c=y_0-3.
\]
Using this value of $c$ we obtain
\[
y(t)=3+(y_0-3) e^{-t/5}.
\]
It yields
\[
\lim_{t\to\infty}y(t)=3=y_{*}.
\]
(b)
\[
y(5)=3-3e^{-1}=1.89636.\qquad\blacksquare \]
\[
y'(t)=\text{salt inflow rate} -\text{salt outflow rate},
\]
we get
\[
y'(t)=0.6-0.2 y(t).
\]
The equilibrium solution is
\[
y_{*}=3. \]
From the differential equation
\[
\frac{1}{0.6-0.2 y}\,dy=1\,dt.
\]
Integrating both sides we obtain
\begin{align*}
&\Step{1A}{-5\ln\left| 0.6-0.2 y\right|=t+c,}\\
&\Step{2A}{\ln\left| 0.6-0.2 y\right|=-\frac{t}{5}+c,} \\
&\Step{3A}{0.6-0.2 y=c e^{-t/5},} \\
&\Step{4A}{y(t)=3+c e^{-t/5}.}\\
\end{align*}
From the initial condition
\[
c=y_0-3.
\]
Using this value of $c$ we obtain
\[
y(t)=3+(y_0-3) e^{-t/5}.
\]
It yields
\[
\lim_{t\to\infty}y(t)=3=y_{*}.
\]
(b)
\[
y(5)=3-3e^{-1}=1.89636.\qquad\blacksquare \]
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