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\title {Solution to the Quiz}

\begin {document}

\maketitle

\noindent We are asked to solve the ODE:

\begin{equation*}
(x^2y^3 - \frac{1}{1 + 9x^2})\frac{dx}{dy} + x^3y^2 = 0
\end{equation*}

\noindent Rearranging the terms, we get:

\begin{equation*}
(x^2y^3 - \frac{1}{1 + 9x^2})dx + x^3y^2dy = 0
\end{equation*}

\noindent And looking at $M(x,y) = x^2y^3 - \frac{1}{1 + 9x^2}$ and
$N(x,y) = x^3y^2$, we find that

\begin{equation*}
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} =
3x^2y^2
\end{equation*}

\noindent so we know we have an exact DE. This means there exists
some $f(x,y)$ such that $\frac{\partial f}{\partial x} = M$ and
$\frac{\partial f}{\partial y} = N$.  Now we have to integrate to
find $f(x,y)$:

\begin{equation} \label{fun}
f(x,y) = \int{M(x,y)}dx + g(y) = \int(x^2y^3 - \frac{1}{1 + 9x^2})dx
+ g(y)
\end{equation}

\noindent To compute the integral $\int{\frac{1}{1 + 9x^2}}dx$, we
need to first use $u$-substitution $u=3x$ (so $du = 3dx$) and then
remember that $\int{\frac{1}{1 + u^2}}du = \arctan(u)$.  Thus

\begin{equation*}
\int{\frac{1}{1 + 9x^2}}dx = \frac{1}{3}\int{\frac{1}{1 + u^2}}du
 = \frac{1}{3}\arctan(u)
 = \frac{1}{3}\arctan(3x)
\end{equation*}

\noindent So after integrating in (\ref{fun}) we get

\begin{equation} \label{fun2}
f(x,y) = \frac{1}{3}x^3y^3 - \frac{1}{3}\arctan(3x) + g(y)
\end{equation}

\noindent Now we take the partial derivative of $f$ with respect to
$y$ and set it equal to $N$:

\begin{equation*}
\frac{\partial f}{\partial y}= x^3y^2 + g'(y)= N = x^3y^2
\end{equation*}

\noindent This tells us that $g'(y) = 0$, so $g(y) = c_1$.  Plugging
back into (\ref{fun2}), we get

\begin{equation*}
f(x,y) = \frac{1}{3}x^3y^3 - \frac{1}{3}\arctan(3x) + c_1
\end{equation*}

\noindent so we have the solution in implicit form when we set
$f(x,y)=0$.  This gives us a correct solution.  However, we can also
multiply through by 3 and set $c = -3c_1$ to get an equivalent,y 
cleaner-looking solution:

\begin{equation*}
x^3y^3 - \arctan(3x) = c
\end{equation*}

\end{document}