}

    \textbf{11.} Nalezn�te v�echna maxim�ln� �e�en� rovnice
    \begin{equation}
        y''' + y'' = 6x + e^{-x}.
    \end{equation}

    \textbf{�e�en�} Provedeme substituci $y'' = u$. Dostaneme rovnici
    \begin{equation}
        u' + u = 6x + e^{-x}.
    \end{equation} Fundament�ln� syst�m t�to rovnice je tvo�en funkc� $e^{-x}$. Partikul�rn� �e�en� zjist�me pomoc� variace
    konstanty:
    \begin{align*}
        u  &= C(x)\cdot e^{-x}\\
        u' &= C'(x)\cdot e^{-x} - C(x)\cdot e^{-x}\\
        u' + u &= C'(x)e^{-x}\\
        &\Downarrow\\
        C'(x)e^{-x} &= 6x + e^{-x}\\
        C'(x) &= 6xe^x + 1\\
        C(x) &= 6e^x(x-1) + x.
    \end{align*}
    �e�en�m rovnice $(2)$ je tedy $u = 6xe^x - 6e^x + x + Ke^{-x}$. �e�en� rovnice $(1)$ dostaneme dvojit�m integrov�n�m:
    \begin{multline*}
        y(x) = \int\left(\int u(x) \dx\right)\dnx =\\
        =\int\left(\int \bigl(6xe^x - 6e^x + x + Ke^{-x}\bigr) \dx \right) \dnx =\\
        =\int\left(6e^x(x-1) - 6e^x + \frac{x^2}{2} - Ke^{-x} + C \right) \dnx =\\
        =6e^x(x-3) + \frac{x^3}{6} + Ke^{-x} + Cx + D.
    \end{multline*}
