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\centerline{MATH 134 -- Calculus with Fundamentals 2}
\centerline{Practice on Separable Differential Equations and Applications}
\centerline{April 18, 2018}
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\noindent
{\it Background}
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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[(2)] 
$$\frac{dy}{dx} = x^3 y^2 + y^2$$ (Hint: factor on the right, then you can separate variables)
\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?
\end{enumerate}
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\caption{The region for problem 4.}
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