自从pgfplots没有“内置”立方根函数，我改编了代码这个问题为了在微分方程斜率场中绘制立方根函数。这个技巧似乎完美地展现了出来：
但是，尽管编译和外观都符合预期，我的 Overleaf 项目返回“缺失数字，视为零”错误。以下是 MWE：
\documentclass{exam}
\printanswers
\usepackage{pgf,tikz}
\usepackage{pgfplots}
\usepackage{physics}
%Slope fields, from https://tex.stackexchange.com/questions/139064/how-to-draw-slope-fields-with-all-the-possible-solution-curves-in-latex
\pgfplotsset{ODE/.style={
axis x line = middle,
axis y line = middle,
xlabel={$x$}, ylabel={$y$},
axis equal image, % Unit vectors for both axes have the same length
view={0}{90}, % We need to use "3D" plots, but we set the view so we look at them from straight up
samples=21, % How many arrows?
cycle list={ % Plot styles
blue,
quiver={
u={1/\length}, v={f(x)/\length}, % End points of the arrows
scale arrows=0.3,
every arrow/.append style={
% -latex % Arrow tip, uncomment to get "arrows" instead of line segments
},
}\\
}
}
}
\begin{document}
\begin{questions}
\question[10]
Consider the differential equation
\begin{equation*}
\dv{y}{x}(x,y)=\frac{x}{y^2},
\end{equation*}
\begin{parts}
\part Sketch the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{center}
\def\length{sqrt(1+(x/y^2)^2)}
\begin{tikzpicture}[declare function={f(\x) = \x/\y^2;}]% Define which function we're using
\begin{axis}[ODE,
width=0.5\textwidth,
xmin=-5, xmax=5,
ymin=-5, ymax=5,
domain=-5:5, y domain=-5:5,
xtick={-5,-4,...,5}, ytick={-5,-4,...,5}
]
\addplot3 (x,y,0);
\addplot[red, samples=50, smooth, thick, no markers, domain=-5:5] {(3/2*x^2-5)/abs(3/2*x^2-5)*abs(3/2*x^2-5)^(1/3)}; %https://tex.stackexchange.com/questions/69411/pgfplots-cant-plot-some-usual-mathematical-functions
\end{axis}
\end{tikzpicture}
\end{center}
\vspace*{\stretch{1}}
\part Solve for the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{solution}
Separating variables,
\begin{align*}
\dv{y}{x}&=\frac{x}{y^2} \\
\int y^2 \dd{y} &= \int x\dd{x} \\
\frac{1}{3}y^3 &= \frac{1}{2} x^2 + C
\end{align*}
Substituting our initial condition,
\begin{align*}
\frac{1}{3} &= 2 + C
\end{align*}
so $C=-\frac{5}{3}$, which implies
\begin{align*}
\frac{1}{3}y^3 &= \frac{1}{2} x^2 - \frac{5}{3} \\
y^3 &= \frac{3}{2} x^2 -5 \\
y &= \sqrt[3]{\frac{3}{2} x^2 -5}
\end{align*}
\end{solution}
\vspace*{\stretch{1}}
\end{parts}
\end{questions}
\end{document}
我发现了一个类似问题另一个人经历过 - 然而解决方案（限制 y 域）似乎在这种情况下没有帮助。
非常感谢您的想法。
编辑：一条评论建议\\从代码的第 22 行删除。这样做可以消除错误，但会导致斜率字段显示不正确：
答案1
有很多问题。其中之一就是除以零。
\documentclass{exam}
\printanswers
\usepackage{pgf,tikz}
\usepackage{pgfplots}
\usepackage{physics}
%Slope fields, from https://tex.stackexchange.com/questions/139064/how-to-draw-slope-fields-with-all-the-possible-solution-curves-in-latex
\pgfplotsset{ODE/.style={
axis x line = middle,
axis y line = middle,
xlabel={$x$}, ylabel={$y$},
axis equal image, % Unit vectors for both axes have the same length
view={0}{90}, % We need to use "3D" plots, but we set the view so we look at them from straight up
samples=21, % How many arrows?
cycle list={ % Plot styles
blue,
quiver={
u={1/\length}, v={f(x,y)/\length}, % End points of the arrows
scale arrows=0.3,
every arrow/.append style={
% -latex % Arrow tip, uncomment to get "arrows" instead of line segments
},
}\\
}
}
}
\begin{document}
\begin{questions}
\question[10]
Consider the differential equation
\begin{equation*}
\dv{y}{x}(x,y)=\frac{x}{y^2},
\end{equation*}
\begin{parts}
\part Sketch the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{center}
\def\length{sqrt(1+(x/y^2)^2)}
\begin{tikzpicture}[declare function={f(\x,\y) = \x/(\y+0.01)^2;}]% Define which function we're using
\begin{axis}[ODE,
width=0.5\textwidth,
xmin=-5, xmax=5,
ymin=-5, ymax=5,
domain=-5:5, y domain=-5:5,
xtick={-5,-4,...,5}, ytick={-5,-4,...,5}
]
\addplot3 (x,y,0);
\addplot[red, samples=50, smooth, thick, no markers, domain=-5:5]
{(3/2*x^2-5)/abs(3/2*x^2-5)*abs(3/2*x^2-5)^(1/3)}; %https://tex.stackexchange.com/questions/69411/pgfplots-cant-plot-some-usual-mathematical-functions
\end{axis}
\end{tikzpicture}
\end{center}
\vspace*{\stretch{1}}
\part Solve for the particular solution to the differential equation satisfying the initial condition $y(2)=1$.
\begin{solution}
Separating variables,
\begin{align*}
\dv{y}{x}&=\frac{x}{y^2} \\
\int y^2 \dd{y} &= \int x\dd{x} \\
\frac{1}{3}y^3 &= \frac{1}{2} x^2 + C
\end{align*}
Substituting our initial condition,
\begin{align*}
\frac{1}{3} &= 2 + C
\end{align*}
so $C=-\frac{5}{3}$, which implies
\begin{align*}
\frac{1}{3}y^3 &= \frac{1}{2} x^2 - \frac{5}{3} \\
y^3 &= \frac{3}{2} x^2 -5 \\
y &= \sqrt[3]{\frac{3}{2} x^2 -5}
\end{align*}
\end{solution}
\vspace*{\stretch{1}}
\end{parts}
\end{questions}
\end{document}