\textbf{8.} Nalezn�te v�echna maxim�ln� �e�en� diferenci�ln� rovnice
    $$
        y' - y^2\cos x = \cos x,
    $$
    kter� vyhovuj� po��te�n� podm�nce $y(0) = 1$.

    \textbf{�e�en�} Uprav�me rovnici na tvar
    $$
        y' = (y^2 + 1) \cos x.
    $$
    Tato rovnice je separovan�, tedy:
    \begin{align*}
        \frac{\dny}{\dnx}       &= (y^2 + 1) \cos x\\
        \frac{\dny}{y^2 + 1}    &= \cos x \dx\\
        \int\frac{\dny}{y^2 + 1}    &= \int\cos x \dx\\
        \arctg y        &= \sin x + C\\
                    &\Downarrow\\
        y           &= \tg(\sin x + C);\\
        \\
        y(0)            &= \tg(C)\\
                    &\Downarrow\\
        C           &= \frac{\pi}{4}.
    \end{align*}
    �e�en�m je tedy $y = \tg\left(\sin x + \frac{\pi}{4}\right)$.
