  \textbf{13.} Nalezn�te v�echna maxim�ln� �e�en� rovnice
    $$
        y''' - y'' + 4y' - 4y = \cos 2x.
    $$

    \textbf{�e�en�} �e��me nejprve homogenn� rovnici
    $$
        y''' - y'' + 4y' - 4y = 0.
    $$
    Jej� charakteristick� polynom je
    $M(\lambda) = \lambda^3 - \lambda^2 + 4\lambda - 4 = (\lambda^2 + 4)(\lambda - 1)$.
    Ko�eny $M(\lambda)$ jsou $\lambda = 1$ a $\mu = \pm 2i$, fundament�ln� syst�m je tedy
    $\lbrace e^x, \sin 2x, \cos 2x \rbrace$.

    Podle v�ty o speci�ln� prav� stran� hled�me partikul�rn� �e�en� ve tvaru $y = Ax\sin 2x + Bx\cos 2x$:
    \begin{align*}
        (-4*) y &= Ax\sin2x + Bx\cos2x\\
        (4*) y' &= 2Ax\cos2x - 2Bx\sin2x + A\sin2x + B\cos2x\\
        (-1*)y''&= -4Ax\sin2x - 4Bx\cos2x + 2A\cos2x - \\
                &- 2B\sin2x + A\cos2x - B\sin2x\\
        y'''    &= -8Ax\cos2x + 8Bx\sin2x - 6A\sin2x - 6B\cos 2x
    \end{align*}
    Tedy
    \begin{multline*}
        -8Ax\cos2x + 8Bx\sin2x - 6A\sin2x - 6B\cos2x+\\
        +4Ax\sin2x + 4Bx\cos2x + 3B\sin2x - 3A\cos2x+\\
        +8Ax\cos2x - 8Bx\sin2x + 4A\sin2x + 4B\cos2x-\\
        -4Ax\sin2x - 4Bx\cos2x =
        (3B - 2A)\sin 2x - (3A + 2B)\cos 2x =
        \cos 2x.
    \end{multline*}
    Tedy mus� b�t $A = \frac{3}{13}$ a $B = \frac{2}{13}$ a �e�en� rovnice m� tvar
    $$
        y = \frac{3x\sin 2x}{13} + \frac{2x\cos 2x}{13} + Ke^x + L\sin 2x + M \cos 2x.
    $$
