<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
\parindent=0pt
\begin{document}

{\bf Question}

\begin{tabular}{llll}
\medskip
1)&amp; $\ds \dy=\frac{x+y}{x-y}$
       &amp; 2)&amp; $\ds \dy=\frac{xy}{x^2+2y^2}\qquad (*)$\\

\medskip
3)&amp; $\ds \dy=\frac{x^2+xy+y^2}{x^2}$\hspace*{2cm}
                &amp; 4)&amp; $\ds \dy=\frac{x^3+3xy^2}{3x^2y+y^3}$\\

\medskip
5)&amp; $\ds x\dy=y+x\cos^2\left(\frac{y}{x}\right)$
                &amp; 6)&amp; $\ds \frac{dx}{dt}=\frac{x}{t}-e^{x/t}\qquad (*)$
\end{tabular}




\vspace{0.25in}

{\bf Answer}

(1-6 are homogeneous)

\begin{itemize}

\item[2)]
$\ds\frac{dy}{dx}=\frac{xy}{x^2+2y^2}=\frac{\frac{y}{x}}{1+2\frac{y}{x}}$

\hspace{1.5in}let $\ds v=\frac{y}{x} \Rightarrow y=xv\Rightarrow
\frac{dy}{dx}=x\frac{dv}{dx}+v$

$\ds x\frac{dv}{dx}+v=\frac{v}{1+2v^2} \Rightarrow
x\frac{dv}{dx}=\frac{-2v^3}{1+2v^2} \Rightarrow
\int\frac{1+2v^2}{2v^3}dv=-\int\frac{1}{x}dx$

$\Rightarrow\ds\int\frac{1}{2v^3}+\frac{1}{v}dv=-\ln x+c
\Rightarrow -\frac{1}{4v^2}+\ln v=-\ln x+c$

$\ds-\frac{x^2}{4y^2}+\ln\left(\frac{y}{x}\right)=-\ln x+c$
\hspace{0.5in} where $\ds-\frac{x^2}{4y^2}+\ln y=c$

\item[4)]
$\ds\frac{dy}{dx}=\frac{x^3+3xy^2}{3x^2y+y^3}
=\frac{1+3\frac{y^2}{x^2}}{3\frac{y}{x}+\frac{y^3}{x^3}}$

\hspace{1.5in} let $\ds v=\frac{y}{x} \Rightarrow
\frac{dy}{dx}=x\frac{dv}{dx}+v$

$\ds x\frac{dv}{dx}+v=\frac{1+3v^2}{3v+v^3} \Rightarrow
x\frac{dv}{dx}=\frac{1-v^4}{3v+v^3} \Rightarrow
\int\frac{3v+v^3}{1-v^4}dv=\int\frac{1}{x}dx$

again use partial fractions.

$\ds\int\frac{v^3}{1-v^4}+\frac{3v}{2(1+v^2)}+\frac{3v}{2(1-v^2)}dv=\ln
x+c$

$\ds-\frac{1}{4}\ln(1-v^4)+\frac{3}{4}\ln(1+v^2)-\frac{3}{4}\ln(1-v^2)=\ln
x+c$

$\ds-\frac{1}{4}\ln[(1-v^2)(1+v^2)]+\frac{3}{4}\ln(1+v^2)-\frac{3}{4}\ln(1-v^2)=\ln
x+c$

$\ds-\frac{1}{4}\ln(1-v^2)-\frac{1}{4}\ln(1+v^2)+\frac{3}{4}\ln(1+v^2)-\frac{3}{4}\ln(1-v^2)=\ln
x+c$

$\ds\frac{1}{2}\ln(1+v^2)-\ln(1-v^2)=\ln x+c$

$\ds\frac{\sqrt{1+v^2}}{1-v^2}=e^cx$

$\ds\frac{\sqrt{1+(\frac{y}{x})^2}}{1-(\frac{y}{x})^2}=Kx$

$\ds\frac{x^2+y^2}{(x^2-y^2)^2}=A$ \hspace{1in}no simplification.

\item[4)]
$\ds\frac{dx}{dt}=\frac{x}{t}-e^\frac{x}{t}$ \hspace{1in}let
$\frac{x}{t}=v \Rightarrow \frac{dx}{dt}=t\frac{dv}{dt}+v$

$\ds\Rightarrow t\frac{dv}{dt}+v=v-e^v \Rightarrow
t\frac{dv}{dt}=-e^v \Rightarrow \int e^{-v}dv=\int\frac{-1}{t}dt$

$\ds\Rightarrow -e^{-v}=-\ln t+c \Rightarrow v=-\ln(\ln t+c)$

$\ds x=-t\ln(\ln t+c)$

\end{itemize}


\end{document}
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