\begin{instructions} In Exercises~\ref{2.nformst}  and~\ref{2.nformend} ,
given the function $\phi$, place the ordinary differential equation
$\phi(t,y,y')=0$ in normal form.
\end{instructions}

\begin{exer}
\begin{exertext}
\label{2.nformst} $\phi(x,y,z)=x^2z+(1+x)y$
\end{exertext}
\begin{soln}
$\phi(t,y,y') = t^2y'+(1+t)y = 0$ must be solved for $y'$.  We get
$$y'= -\frac{(1+t)y}{t^2}.$$
\end{soln}
\begin{answer}$y'= -\dfrac{(1+t)y}{t^2}.$\end{answer}
\end{exer}

\begin{exer}
\begin{exertext}
\label{2.nformend} $\phi(x,y,z)=xz-2y-x^2$
 \end{exertext}
\begin{soln}
 $\phi(t,y,y')=ty'-2y-t^2$ must be solved for $y'$. We get
  $$y'=\frac{2y+t^2}{t}.$$
 \end{soln}
\end{exer}

\begin{instructions}
In Exercises~\ref{2.1st}--\ref{exer2.1.6}, show that the given
solution is a general
solution of the differential equation. Use a computer or calculator to
sketch the solutions for the given values of the
arbitrary constant. Experiment with different intervals for $t$ until you
have a plot that shows what you consider to be the most important
behavior of the family.
\end{instructions}

\begin{exer}
\begin{exertext}
\label{2.1st} $y'=-ty$, $y(t)=Ce^{-(1/2)\,t^2}$, $C=-3,-2,\ldots,3$
\end{exertext}
\begin{soln}
$y(t)' = -Cte^{-(1/2)\,t^2}$ and $-ty(t) = -tCe^{-(1/2)\,t^2}$, so
$y'=-ty$. \par
\exfig{ex2.1ans01}
\end{soln}
\begin{answer}\exfig{ex2.1ans01}\end{answer}
\end{exer}