�e�en� mus� b�t kladn� ({$ y>0 $}), jinak nem� rovnice smysl. Za tohoto
p�edpokladu pou�ijeme metodu separace prom�nn�ch.
\item{i}
Funkce  {$ y=1, x\in (0,\pi ) $} je trivi�ln�m �e�en�m, a
vzhledem k~tomu, �e defini�n� interval vzhledem k~zad�n� nelze ji�
roz���it, je i maxim�ln�m �e�en�m. Toto �e�en� z�ejm� vyhovuje
zad�n�, tak�e jsme zodpov�d�li bod a).
\item{ii}
 Proch�z�-li bodem {$ (\frac{\pi }{2},1) $} je�t� jin� �e�en�,
mus� toto �e�en� b�t pro n�jak� {$ x\in (0,\pi )\ ( x\neq \frac{\pi
}{2}) $} r�zn� od jedn�.
Hledejme tedy �e�en�, proch�zej�c�  bodem {$ (x, y),\allowmathbreak x\in (0,\pi ),
0<y\neq 1.  $}
 Jestli�e ��dn� takov� �e�en� nenab�v� nikde (nejen v~bod� {$
 x=\frac{\pi }{2} $}) hodnotu jedna (ani ve
 smyslu limitn�m), je \cuvo maxim�ln� trivi�ln�
 �e�en�'' jedin�m �e�en�m �lohy; naopak  �e�en�, kter� se v~n�jak�m bod�
 rovn� jedn�, lze
 \cuvo nalepit" na trivi�ln� �e�en� a jednozna�nost tak poru�it.
 \itemitem{} Pro hledan� �e�en� mus� platit (separace):
 $$
   \align
      y'\sin x& = y\ln y\cr
      \frac{y'}{y}\frac{1}{\ln y}&=\frac{1}{\sin x}\cr
      \int_{}^{}\frac{y'}{y}\frac{1}{\ln y}\dx&=\int_{}^{}\frac{1}{\sin
      x}\dx\cr
      \intertext{substituce {$ y(x)=y, y'\dx=\dy $}:}
      \int_{}^{}\frac{\dy}{y}\frac{1}{\ln y}&=\int_{}^{}\frac{1}{\sin
      x}\dx\cr
      \intertext{substituce {$ \ln y =z, \frac{\dy}{y}=\dz $}:}
      \int_{}^{}\frac{\dz}{z}&=\int_{}^{}\frac{1}{\sin
      x}\dx\cr
          \ln |z|=\ln|\ln y|&=\int_{}^{}\frac{\dx}{\sin x} \cr
   \endalign
$$
Integr�l {$ \int_{}^{}\frac{\dx}{\sin x} $}, (kter� m� na zadan�m
intervalu {$ x\in (0,\pi ) $} smysl), lze spo��tat r�zn�mi zp�soby:
\item{($\alpha$) }Integrand roz����me v�razem {$ \sin x $} a u�ijeme
substituci {$ \cos x = z, \sin x \dx = -\dz $}:
$$
  \multline
  \int_{}^{}\frac{\dx}{\sin x}=\cr
  =\int_{}^{}\frac{\sin x}{1-\cos ^{2}x}
  =\int_{}^{}  \frac{-\dz}{1-z^{2}}=-\frac{1}{2}\int_{}^{}\left(
  \frac{1}{1+z}+\frac{1}{1-z}\right)\dz=\cr
  =\ln c\sqrt{\frac{1-\cos x}{1+\cos x}}, \quad c=e^{k}>0
  \endmultline
$$
(zde {$ k\in \er $} je integra�n� konstanta,) odkud
\smallskip
\centerline{$| \ln y |= c\sqrt{\frac{1-\cos x}{1+\cos x}}. $}

Existuje-li netrivi�ln� �e�en� dan� po��te�n� �lohy (tj. rovnice 
s~podm�nkou \newline{$ y(\frac{\pi }{2})=1 $}), mus� existovat takov� konstanta {$
c$} a takov� {$ x_0\in (0,\pi ) $} (nikoliv nutn� {$ x_{0}=\frac{\pi
}{2} $}),
�e plat� {$ y(x_{0})=1, \ln y = c\sqrt{\frac{1-\cos x}{1+\cos x}} =0$}.
\par
Vid�me ov�em, �e v�dy plat�  {$ \sqrt{\frac{1-\cos x}{1+\cos x}} \neq 0
$},
tak�e odsud plyne {$ c= 0, $} co� je ov�em zak�zan� hodnota konstanty
{$ c $} a �e�en� jdouc� bodem {$ (\frac{\pi }{2},1) $} je tedy ur�eno
jednozna�n�.
\par
(Rozmyslete si, co se stane, jestli�e �e��me na�i rovnici na intervalu
{$ (-\pi, \pi ) $} s~podm�nkou {$ y(0)=1. $})
\item{({$ \beta  $})}Paradoxn� skoro jednodu��� je substituce
{$\tg\frac{x}{2}=u. $} Pak
$$
  \alignat2
         \sin x&=   & 2\sin \frac{x}{2} \cos \frac{x}{2} &=\frac{2u}{1+u^{2}}\cr
         \cos x&=   &\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2} &=\frac{1-u^{2}}{1+u^{2}}\cr
  \frac{1}{2}\frac{\dx}{\cos^{2}\frac{x}{2}}&=\dt&&\cr
    \dx&=   & 2\cos^{2}         \frac{x}{2} \dt&= \frac{2\du}{1+u^{2}}
  \endalignat
$$
(tato substituce je pou�iteln� nejen na intervalu {$ (0,\pi) $}, ale i
na {$(-\pi ,\pi );  $} v~bod� {$ x=0 $} ov�em ztr�c� smysl integrand),
\par a tedy
  $$
  \int_{}^{}\frac{\dx}{\sin x}=\int_{}^{}\frac{1+u^{2}}{2u}
  \frac{2u}{1+u^{2}} \du=\int_{}^{}\frac{\du}{u}= \ln| u| +k = \ln |c\tg
  \frac{x}{2}|;
$$
dal�� postup je analogick� jako v~bod� {$ (\alpha).  $}
