\documentclass{article}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}

\usepackage{amsmath}

\pgfplotsset{ % Define a common style
    SField/.style={
        width=0.7\textwidth, % Overall width of the plot
        axis equal image, % Unit vectors for both axes have the same length
        view={0}{90}, % We need to use "3D" plots; view is from straight up
        xmin=-10, xmax=10, % Axis limits
        ymin=-5, ymax=5,
				zmin=0.0, zmax = 0.1,
				domain=-10:10, y domain=-5:5, % Domain of function
        xtick={-10,-5,...,10}, ytick={-4,-2,...,4}, % Tick marks
				%axis lines = center,
        samples = 21, samples y = 11,% How many arrows?
    }
}

\begin{document}
Match the slope field with its description:



\begin{tikzpicture}[declare function={f(\x,\y) = -\x/\y;}]
\begin{axis}[SField]
\def\VScale{.5} %scale of field lines
\def\Vlen{sqrt(1 + f(x,y)^2)}
\addplot3[gray,
					samples y = 11,
					domain=-10.001:10, y domain=-5.001:5, % Domain; .0001 is to avoid /0
					quiver={u ={0.5/\Vlen}, v={0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
\addplot3[gray,
					samples y = 11,
					domain=-10.001:10, y domain=-5.001:5, % Domain of function
					quiver={u={-0.5/\Vlen}, v={-0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
%\addplot[]{0};
%\addplot[]({0},{\x});
\fill[color=white] (axis cs:0,0) circle (.5);
\end{axis}
\end{tikzpicture}

The solution curves satisfy the differential equation $\frac{dy}{dx}  = -\frac{x}{y}$


\begin{tikzpicture}[declare function={f(\x,\y) = \x+\y;}]
\begin{axis}[SField]
\def\VScale{.5} %scale of field lines
\def\Vlen{sqrt(1 + f(x,y)^2)}
\addplot3[gray,
					samples y = 11,
					domain=-10:10, y domain=-5.001:5, % Domain; .0001 is to avoid /0
					quiver={u ={0.5/\Vlen}, v={0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
\addplot3[gray,
					samples y = 11,
					domain=-10:10, y domain=-5.001:5, % Domain of function
					quiver={u={-0.5/\Vlen}, v={-0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
%\addplot[]{0};
%\addplot[]({0},{\x});
\end{axis}
\end{tikzpicture}

The solution curves satisfy the differential equation $\frac{dy}{dx}  = x + y$

\begin{tikzpicture}[declare function={f(\x,\y) = (0.1)*(\y-2)*\x;}]
\begin{axis}[SField]
\def\VScale{.5} %scale of field lines
\def\Vlen{sqrt(1 + f(x,y)^2)}
\addplot3[gray,
					samples y = 11,
					domain=-10:10, y domain=-5.001:5, % Domain; .0001 is to avoid /0
					quiver={u ={0.5/\Vlen}, v={0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
\addplot3[gray,
					samples y = 11,
					domain=-10:10, y domain=-5.001:5, % Domain of function
					quiver={u={-0.5/\Vlen}, v={-0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
%\addplot[]{0};
%\addplot[]({0},{\x});
\end{axis}
\end{tikzpicture}

There is exactly one equilibrium solution.%, and it is unstable.


\begin{tikzpicture}[declare function={f(\x,\y) = 0.25*(\y+2)*(2-\y);}]
\begin{axis}[SField]
\def\VScale{.5} %scale of field lines
\def\Vlen{sqrt(1 + f(x,y)^2)}
\addplot3[gray,
					samples y = 11,
					domain=-10:10, y domain=-5.001:5, % Domain; .0001 is to avoid /0
					quiver={u ={0.5/\Vlen}, v={0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
\addplot3[gray,
					samples y = 11,
					domain=-10:10, y domain=-5.001:5, % Domain of function
					quiver={u={-0.5/\Vlen}, v={-0.5*f(x,y)/\Vlen}, % End points of the arrows
									scale arrows=\VScale,
									every arrow/.append style={-}, % Arrow tip
          }](x,y,0);
%\addplot[]{0};
%\addplot[]({0},{\x});
\end{axis}
\end{tikzpicture}

%For $0 \leq y \leq 3$, the solution curves are logistic.
The solution curve $y=f(x)$ that goes through $(0,0)$ satisfies $\displaystyle \lim_{x \to \infty}f(x) = 2$.


\end{document} 