一个盒子套着另一个盒子

Question

我尝试将一个框（红色框）放入另一个框（黄色框）内，但是这样做时红色框会失去其颜色格式。

这是我的代码：

\begin{ejemplo}
    
        $ {{e^{x}=(1+e^{x})yy^{\prime}}}$\\
    
    \textbf{PASO 1.} Ordenando en la forma general $ M(x,y)dx+N(x,y)dy=0 $.
    \begin{equation*}
        e^{x}-(1+e^{x})y\dfrac{dy}{dx}=0
    \end{equation*}
    \begin{equation*}%
        e^{x}dx-(1+e^{x})y{dy}=0
    \end{equation*}
    \textbf{PASO 2.} Identificando el tipo de ecuación
    \begin{equation*}
        \dfrac{e^{x}}{\left(1+e^{x}\right)}dx-ydy=0\quad m/m\int\quad\Rightarrow \text{E.D.O. Variables separables}
    \end{equation*} 
    
\begin{minipage}{10cm}
    \textbf{PASO 3.} resolviendo la E.D.O.  
    \begin{equation*}%
        \dfrac{e^{x}}{\left(1+e^{x}\right)}dx-y{dy}=0\: [m/m]\int
    \end{equation*}
    
    \begin{equation*}%
        \int\dfrac{e^{x}}{\left(1+e^{x}\right)}dx-\int y{dy}=\int 0
    \end{equation*}
    \begin{center}
        \fcolorbox{yellow}{yellow}{ \color{black} $ \ln\left(1+e^{x}\right)-\dfrac{y^{2}}{2}=c$}
    \end{center}
\end{minipage}
\begin{minipage}{5cm}   
    \tcbset{
        enhanced,
        colback=red!5!white,
        boxrule=0.1pt,
        colframe=red!75!black,
        fonttitle=\bfseries
    }
    
    \begin{tcolorbox}[title=Cálculo auxiliar,center title,hbox,    %%<<---- here
        lifted shadow={1mm}{-2mm}{3mm}{0.1mm}%
        {black!50!white}]
        \begin{varwidth}{\textwidth}
            \begin{equation*}
                \int\dfrac{e^{x}}{\left(1+e^{x}\right)} =\int\dfrac{dt}{t}
            \end{equation*}
            
            \begin{equation*}
                \left\{ \begin{array}{rc}
                    t &=1+e^{x} \\
                    dt & =e^{x}dx \\
                \end{array}\right.=\ln\left(1+e^{x} \right)
            \end{equation*}     
        \end{varwidth}
    \end{tcolorbox}
\end{minipage}
\end{ejemplo}

Answer 1

您没有提供完整的 MWE（ejemplo环境如何定义？），因此很难回答。

我希望这有帮助。

\documentclass{article}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{array}
\usepackage{varwidth}
\usepackage[all]{tcolorbox}
\newtcolorbox{ejemplo}[2][]{enhanced,
attach boxed title to top left={yshift*=3mm-\tcboxedtitleheight},
boxed title style={fonttitle=\bfseries},
breakable,
title={#2},#1}

\newtcolorbox{boxinbox}[2][]{
        enhanced,
        colback=red!5!white,
        boxrule=0.1pt,
        colframe=red!75!black,
        fonttitle=\bfseries,
        center title,
        width=.4\linewidth,
        nobeforeafter,
        lifted shadow={1mm}{-2mm}{3mm}{0.1mm}%
        {black!50!white},
        title={#2},
        #1
    }
\begin{document}
\begin{ejemplo}{Ejemplo}
    
        $ {{e^{x}=(1+e^{x})yy^{\prime}}}$\newline
    
    \textbf{PASO 1.} Ordenando en la forma general $ M(x,y)dx+N(x,y)dy=0 $.
    \begin{equation*}
        e^{x}-(1+e^{x})y\dfrac{dy}{dx}=0
    \end{equation*}
    \begin{equation*}%
        e^{x}dx-(1+e^{x})y{dy}=0
    \end{equation*}
    \textbf{PASO 2.} Identificando el tipo de ecuación
    \begin{equation*}
        \dfrac{e^{x}}{\left(1+e^{x}\right)}dx-ydy=0\quad m/m\int\quad\Rightarrow \text{E.D.O. Variables separables}
    \end{equation*} 
    \textbf{PASO 3.} resolviendo la E.D.O.
\begin{center}    
    \begin{minipage}[b]{.4\linewidth}
    \begin{equation*}%
        \dfrac{e^{x}}{\left(1+e^{x}\right)}dx-y{dy}=0\: [m/m]\int
    \end{equation*}
    \begin{equation*}%
        \int\dfrac{e^{x}}{\left(1+e^{x}\right)}dx-\int y{dy}=\int 0
    \end{equation*}
    \begin{center}
        \fcolorbox{yellow}{yellow}{ \color{black} $ \ln\left(1+e^{x}\right)-\dfrac{y^{2}}{2}=c$}
    \end{center}
    \end{minipage}
\begin{boxinbox}{Cálculo auxiliar}
        \begin{equation*}
    \int\dfrac{e^{x}}{(1+e^{x})} =\int\dfrac{dt}{t}
            \end{equation*}
            \begin{equation*}
                \left\{ \begin{array}{rc}
                    t &=1+e^{x} \\
                    dt &=e^{x}dx \\    \end{array}\right. =\ln(1+e^{x})
            \end{equation*}    
\end{boxinbox}
\end{center}
\end{ejemplo}
\end{document}