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\centerline{MATH 136 -- Calculus 2}
\centerline{Practice on Separable Differential Equations and Applications}
\centerline{March 20, 2018}
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\noindent
{\it Background}
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A differential equation is a relation between derivative(s) of an unknown 
function and other known functions.  A differential equation often serves
as a \emph{mathematical model} for how some quantity is evolving with respect to time.
For example, Newton's Law of Cooling (and Heating) says: \emph{if an object is
placed into a surrounding medium held at constant temperature $A$, then the 
object's temperature changes at a rate proportional to the difference between
its temperature and the ambient temperature}.  If $T$ is the object's temperature,
this statement is equivalent to the differential equation
$$\frac{dT}{dt} = k(T - A)$$
for some constant $k$.  

Any differential equation of the form 
$$\frac{dy}{dx} = g(y)\cdot h(x)$$
is called a ``separable'' equation and can be solved by 
\begin{itemize}
\item {\it separating the variables} to the form
$$\frac{dy}{g(y)} = h(x)\ dx$$
\item {\it integrating on both sides} (on the left, treat the variable of 
integration as $y$, not $x$; this can be justified by the substitution method
for integration):
$$\int\frac{dy}{g(y)} = \int h(x)\ dx$$
\item {\it solving the resulting equation for $y$}
\end{itemize} 
\begin{enumerate}
\item[(A)]  To practice, solve the following separable equations:
\begin{enumerate}
\item[(1)]
$$\frac{dy}{dx} = \frac{y}{x^2 + 1}$$
\item[] {\it Solution:}  Separated:
$$\frac{dy}{y} = \frac{1}{x^2 + 1}\ dx$$
We integrate both sides:
$$\int \frac{dy}{y} = \int \frac{1}{x^2 + 1}\ dx \Rightarrow \ln|y| = \tan^{-1}(x) + C$$
To solve for $y$, we exponentiate, then take both choices of sign:
$$y = \pm e^C e^{\tan^{-1}(x)} = k e^{\tan^{-1}(x)}$$
where $k$ is a positive or negative constant: $k = \pm e^C$.   
\item[(2)] 
$$\frac{dy}{dx} = x^3 y^2 + y^2$$ (Hint: factor on the right, then you can separate variables)
\item[] {\it Solution:} Factored,
$\frac{dy}{dx} = y^2(x^3 + 1)$, so 
$$\int\frac{dy}{y^2} = \int x^3 + 1\ dx$$
and hence
$$\frac{-1}{y} = \frac{x^4}{4} + x + C$$
Then solving for $y$, 
$$y = \frac{-1}{x^4/4 + x + C}$$

\end{enumerate}
\item[(B)]  Solve the following Newton's Law of Cooling problem:  A hot cup of 
coffee is poured at time $t = 0$ with the temperature being $80^\circ$ C.  The
cup is placed on a desk in a room with temperature maintained at $23^\circ$ C.  Five minutes
later, the coffee has cooled to $70^\circ$ C.  At what time will the coffee
have cooled down to $40^\circ$ F?
\item[] {\it Solution:}  Here the ambient temperature is $A = 23$ degrees C.
From the explanation above, we know the temperature $T$ of the cup of coffee satisfies
$$\frac{dT}{dt} = k(T - 23)$$
This separable: 
$$\int \frac{dT}{T - 23} = \int k dt$$
so 
$$\ln|T - 23| = k t + C$$
and 
$$T = 23 + D e^{kt}$$
for some positive or negative constant $D = \pm e^C$.  When $t = 0$ we know $T = 80$, 
so $80 = 23 + D e^{k\cdot 0} = 23 + D$.  Hence $D = 57$.  Then when $t = 5$, we have
$70 = 23 + 57 e^{5k}$, so $e^{5k} = \frac{47}{57}$ and $k = \frac{\ln(47/57)}{5} \doteq -.0386$.
Finally we need to solve the equation 
$$40 = 23 + 57 e^{(-.0386)t}$$
for $t$.  The solution is 
$$t = \frac{\ln(17/57)}{-.0386} \doteq 31.4\text{ minutes}$$
\end{enumerate}
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\begin{figure}
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\includegraphics[width=.8\linewidth]{needleregion.png}
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\caption{The region for problem 4.}
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