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

{\bf Question}

The following question is adapted from R.A.Adams "Calculus"
exercises 19.2

Solve the following differential equations\\
\begin{tabular}{llll}

\medskip
1) & $\ds\dy=\frac{y}{2x}$      & 2) & $\ds
\dy=\frac{3y-1}{x}\qquad (*)$\\

\medskip
3) & $\ds \dy =\frac{x^2}{y^2}$ & 4) & $\ds \dy=x^2y^2$\\

\medskip
5) & $\ds \dy=\frac{x^2}{y^3}$  & 6) & $\ds \dy=x^3y^4\qquad
(*)$\\

\medskip
7) & $\ds \frac{dY}{dt}=tY$              & 8) &
$\ds\frac{dx}{dt}=e^x \cos t$\\

\medskip
9) & $\ds \dy=1-y^2$            & 10)& $\ds \dy=1+y^2$\\

\medskip
11)& $\ds\frac{dy}{dt}=2+e^y$           & 12)& $\ds
\dy=y^2(1-y)\qquad (*)$\\

\medskip
13)& $\ds \dy=\sin x \cos^2 y$  & 14)& $\ds x\dy=y\ln x$
\end{tabular}




\vspace{0.25in}

{\bf Answer}

(odd solutions in Adams \lq\lq Calculus\rq\rq)

(1-14 are separable)
\begin{itemize}
\item[2)]
$\ds\frac{dy}{dx}=\frac{3y-1}{x} \Rightarrow
\int\frac{dy}{3y-1}=\int\frac{dx}{x} \Rightarrow
\frac{1}{3}\ln(3y-1)=\ln x+c$

$\ds(3y-1)^\frac{1}{3}=e^cx \Rightarrow 3y-1=kx^3 \Rightarrow
y=Ax^3+\frac{1}{3}$

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

$\ds y=\frac{1}{A-\frac{1}{3}x^3}$

\item[6)]
$\ds\frac{dy}{dx}=x^3y^4 \Rightarrow \int\frac{dy}{y^4}=\int x^3dx
\Rightarrow -\frac{1}{3y^3}=\frac{1}{4}x^4+c$

$\ds y^3=\frac{1}{A-\frac{3}{4}x^4} \Rightarrow
y=\left(\frac{1}{A-\frac{3}{4}x^4}\right)^\frac{1}{3}$


\item[8)]
$\ds\frac{dx}{dt}=e^x\cos t \Rightarrow \int e^{-x}dx=\int\cos tdt
\Rightarrow -e^{-x}=\sin t+c$

$\ds e^{-x}=-\sin t-c \Rightarrow x=-\ln(-\sin t-c)$

\item[10)]
$\ds\frac{dy}{dx}=1+y^2 \Rightarrow \int\frac{dy}{1+y^2}=\int dx
\Rightarrow \tan^{-1}y=x+c$

$\ds y=\tan(x+c)$

\item[12)]
$\ds\frac{dy}{dx}=y^2(1-y) \Rightarrow
\int\frac{dy}{y^2(1-y)}=\int dx$ then partial fractions

$\ds\int\frac{1}{y^2}+\frac{1}{y}+\frac{1}{1-y}dy=x+c \Rightarrow
-\frac{1}{y}+\ln y-\ln(1-y)=x+1$

can't simplify.

\item[14)]
$\ds x\frac{dy}{dx}=y\ln x \Rightarrow
\int\frac{1}{y}dy=\int\frac{\ln x}{x}dx$ (use substitution $\ln
x=q$ to do integration)

$\ds\ln y=\frac{1}{2}(\ln x)^2+c \Rightarrow
y=e^ce^{\frac{1}{2}(\ln x)^2)}$
\end{itemize}



\end{document}
