Пакет упражнений - Короткие ответы и длинные ответы

Пакет упражнений - Короткие ответы и длинные ответы

Можно ли включить в пакет упражнений как краткие, так и развернутые ответы?

Например, в следующем примере:Math-Book-как-писать-упражнения-и-ответы - обмен стеками

Возможно ли сделать раздел с краткими ответами, содержащими только ответы, а затем полностью проработанный раздел с развернутыми ответами?

МВЭ:

\documentclass[11pt]{article}
\usepackage[margin=2cm,includefoot,bottom=2.55cm,top=2.025cm,headsep=0.5cm,footskip=0.65cm]{geometry}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{multicol}

\usepackage{ifthen}
\newboolean{firstanswerofthesection}  

\usepackage{xcolor}
\definecolor{e}{RGB}{0,40,120}

\usepackage{chngcntr}
\usepackage{stackengine}

\usepackage{tasks}
\newlength{\longestlabel}
\settowidth{\longestlabel}{(m)}
\settasks{label-format=\color{e}, counter-format={(tsk[a])}, label-width=\longestlabel,
    item-indent=0pt, label-offset=2pt, column-sep={10pt}}

\usepackage[lastexercise,answerdelayed]{exercise}
\counterwithin{Exercise}{section}
\counterwithin{Answer}{section}
\renewcounter{Exercise}[section]
\newcommand{\QuestionNB}{\color{e}\bfseries\arabic{Question}.\,}
\renewcommand{\ExerciseName}{EXERCISE}
\renewcommand{\ExerciseHeader}{\noindent\def\stackalignment{l}% code from https://tex.stackexchange.com/a/195118/101651
    \stackunder[0pt]{\colorbox{e}{\textcolor{white}{\textbf{\large\ExerciseName\;\large\ExerciseHeaderNB}}}}{\textcolor{e}{\rule{\linewidth}{2pt}}}\medskip}
\renewcommand{\AnswerName}{Exercises}
\renewcommand{\AnswerHeader}{\ifthenelse{\boolean{firstanswerofthesection}}%
    {\bigskip\noindent\textcolor{e}{\textbf{SECTION \thesection}}\newline\newline%
        \noindent\bfseries\emph{\textcolor{e}{\AnswerName\ \ExerciseHeaderNB, page %
                \pageref{\AnswerRef}}}\smallskip}
    {\noindent\bfseries\emph{\textcolor{e}{\AnswerName\ \ExerciseHeaderNB, page \pageref{\AnswerRef}}}\smallskip}}
\setlength{\QuestionIndent}{16pt}

\begin{document}
    \section{First}

    \begin{Exercise}\label{EX11}
        \vspace{-\baselineskip}% <-- You don't need this line of code if there's some text here
        \Question In problem \ref{EX11-1-i}-\ref{EX11-1-iii}, determine whether the given differential equation is separable  
        \begin{tasks}(2)
            \task\label{EX11-1-i} $\frac{dy}{dx}-\sin{(x+y)}=0$     
            \task $\frac{dy}{dx}=4y^2-3y+1$ 
            \task\label{EX11-1-iii} $\frac{ds}{dt}=t\ln{(s^{2t})}+8t^2$ 
        \end{tasks}
        \Question In problem \ref{EX11-2-iv}-\ref{EX11-2-viii}, solve the equation 
        \begin{tasks}[resume=true](2)
            \task\label{EX11-2-iv} $\frac{dx}{dt}=3xt^2$
            \task $y^{-1}dy+ye^{\cos{x}}\sin{x}dx=0$
            \task $(x+xy^2)dx+ye^{\cos{x}}\sin{x}dx=0$
            \task\label{EX11-2-viii} $\frac{dy}{dt} = \frac{y}{t+1} + 4t^2 +  4t$, $\quad$ $y(1) = 10$
        \end{tasks}
    \end{Exercise}
    \setboolean{firstanswerofthesection}{true}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX11}]
            \Question 
            \begin{tasks}(3)
                \task 45
                \task 32
                \task 32
            \end{tasks} 
            \Question 
            \begin{tasks}[resume=true]
                \task This is a solution of Ex 4
                \task This is a solution of Ex 5 
                \task This is a solution of Ex 6 
                \task This is a solution of Ex 7 
            \end{tasks} 
        \end{Answer}
    \end{multicols}
    \setboolean{firstanswerofthesection}{false}

    \begin{Exercise}\label{EX12}
        Another exercise. 
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use
        \Question If you don't need a horizontal list, you can simply use

    \end{Exercise}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX12}]
            \Question This is a solution of Ex 1
        \end{Answer}
    \end{multicols}

    \section{Second}

    \begin{Exercise}\label{EX21}
        \vspace{-\baselineskip}% <-- You don't need this line of code if there's some text here
        \Question Eight systems of differential equations and five direction fields are given below.  Determine the system that corresponds to each direction field and sketch the solution curves that correspond to the initial conditions $(x_0, y_0) = (0,1)$ and $(x_0, y_0) = (1,-1)$.
        \begin{tasks}(3)
            \task $\begin{aligned}
            \frac{dx}{dt} & = -x \\     
            \frac{dy}{dt} & = y-1
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = x^2 - 1 \\        
            \frac{dy}{dt} & = y
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = x+2y \\
            \frac{dy}{dt} & = -y
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = 2x \\
            \frac{dy}{dt} & =  y
            \end{aligned}$
            \task $\begin{aligned}
            \frac{dx}{dt} & = x \\
            \frac{dy}{dt}  & = 2y
            \end{aligned}$ 
            \task$\begin{aligned}
            \frac{dx}{dt} & = x-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}$
            \task$\begin{aligned}
            \frac{dx}{dt} & = x^2-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}$        
            \task $\begin{aligned}
            \frac{dx}{dt} & = x- 2y \\
            \frac{dy}{dt} & =  -y
            \end{aligned}$
        \end{tasks}
    \end{Exercise}
    \setboolean{firstanswerofthesection}{true}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX21}]
            \Question 
            \begin{tasks}
                \task This is a solution of Ex 1
                \task This is a solution of Ex 2 
                \task This is a solution of Ex 3 
                \task This is a solution of Ex 4 
                \task This is a solution of Ex 5 
                \task This is a solution of Ex 6 
                \task This is a solution of Ex 7 
                \task This is a solution of Ex 8 
            \end{tasks}
        \end{Answer}
    \end{multicols}
    \setboolean{firstanswerofthesection}{false}
    \newpage        
    \begin{Exercise}\label{EX22}
        Since these are systems, maybe it's better to put the \verb|aligned| enviroment within  \verb|\left\{| and \verb|\right.|: 
        \Question Eight systems of differential equations and five direction fields are given below.  Determine the system that corresponds to each direction field and sketch the solution curves that correspond to the initial conditions $(x_0, y_0) = (0,1)$ and $(x_0, y_0) = (1,-1)$.
        \begin{tasks}(3)
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = -x \\     
            \frac{dy}{dt} & = y-1
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x^2 - 1 \\        
            \frac{dy}{dt} & = y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x+2y \\
            \frac{dy}{dt} & = -y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = 2x \\
            \frac{dy}{dt} & =  y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x \\
            \frac{dy}{dt}  & = 2y
            \end{aligned}\right.$ 
            \task$\left\{\begin{aligned}
            \frac{dx}{dt} & = x-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}\right.$
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x^2-1 \\
            \frac{dy}{dt} & = -y
            \end{aligned}\right.$        
            \task $\left\{\begin{aligned}
            \frac{dx}{dt} & = x- 2y \\
            \frac{dy}{dt} & =  -y
            \end{aligned}\right.$
        \end{tasks}
    \end{Exercise}
    \begin{multicols}{2}
        \begin{Answer}[ref={EX22}]
            \Question 
            \begin{tasks}
                \task This is a solution of Ex 1
                \task This is a solution of Ex 2 
                \task This is a solution of Ex 3 
                \task This is a solution of Ex 4 
                \task This is a solution of Ex 5 
                \task This is a solution of Ex 6 
                \task This is a solution of Ex 7 
                \task This is a solution of Ex 8 
            \end{tasks}
        \end{Answer}
    \end{multicols}

    \newpage

    \section{Answer to all problems}

    \begin{multicols}{2}\raggedcolumns
        \shipoutAnswer
    \end{multicols}

\end{document}

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