\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
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\begin{document}

{\bf Question}

Solve the following Bernoulli equations
\begin{enumerate}
\item
$$ x \dy = y - {1 \over y} \qquad (*) $$
\item
$$ 3 \dy + x y = \frac{x}{y^2} \qquad (*) $$

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds x\frac{dy}{dx}-y=-\frac{1}{y}$ is a Bernoulli equation with
$\ds n=-1$, so put $\ds y=u^\frac{1}{1-n}=u^\frac{1}{2}
\Rightarrow$ either $\ds u^\frac{-1}{2}=0$ or
$\ds\frac{1}{2}x\frac{du}{dx}-u=-1$

$\ds\frac{du}{dx}-\frac{2}{x}u=-\frac{2}{x} \Rightarrow
I(x)=exp\left(\int -\frac{2}{x}dx\right)=exp(-2\ln
x)=\frac{1}{x^2}$

$\ds \frac{d}{dx}\left(\frac{1}{x^2}u\right)=-\frac{2}{x^3}
\Rightarrow \frac{1}{x^2}u=\frac{1}{x^2}+A \Rightarrow u=1+Ax^2$

$\ds y=\left(1+Ax^2\right)^\frac{1}{2}$

\item[b)]
$\ds 3\frac{dy}{dx}+xy=\frac{x}{y^2}$ is a Bernoulli equation with
$\ds y=-2$

put $\ds y=u^\frac{1}{1-n}=u^\frac{1}{3} \Rightarrow$ either $\ds
u^\frac{-2}{3}=0$ or $\ds\frac{du}{dx}+xu=x$

So $\ds I(x)=exp\left(\int xdx\right)=e^{\frac{1}{2}x^2}$

$\ds\frac{d}{dx}\left(e^{\frac{1}{2}x^2}u\right)=xe^{\frac{1}{2}x^2}
\Rightarrow e^{\frac{1}{2}x^2}u=e^{\frac{1}{2}x^2}+A$

$\ds u=1+Ae^{-\frac{1}{2}x^2} \Rightarrow
y=\left(1+Ae^{-\frac{1}{2}x^2}\right)^\frac{1}{3}$

\end{itemize}


\end{document}
