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\begin{document}

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%   ____      _                    _ _           
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{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 2
\par PURO-UFF - 2015.2
\par P2 - 21/mar/2016 - Eduardo Ochs
% \par Versão: 14/mar/2016
% \par Links importantes:
% \par \url{http://angg.twu.net/2015.2-C2.html} (página do curso)
% \par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros)
% \par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
% \par {\tt eduardoochs@gmail.com} (meu e-mail)
}

\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B       (#1 pts){{\bf(#1 pts)}}

\bsk
\bsk

1) \T(Total: 2.5 pts) Seja $(*)$ esta EDO: $f'' + 5f' + 6f = 0$.

a) \B(1.0 pts) Encontre as soluções básicas de $(*)$.

b) \B(0.5 pts) Encontre uma solução de $(*)$ que obedeça $f(0)=0$ e $f(1)=1$.

c) \B(0.5 pts) Encontre uma solução de $(*)$ que obedeça $f(0)=1$ e $f(1)=0$.

d) \B(0.5 pts) Encontre uma solução de $(*)$ que obedeça $f(0)=2$ e $f(1)=3$.


\bsk

2) \T(Total: 2.5 pts) Seja $(**)$ esta EDO: $(D-(a+ib))(D-(a-ib))f =
0$ (obs: $a,b∈\R$).

a) \B(1.0 pts) Encontre as soluções básicas de $(**)$.

b) \B(1.0 pts) Encontre $a$ e $b$ para que $f(x) = e^{-x} \cos 2x$
seja solução de $(**)$.

c) \B(0.5 pts) Com o $a$ e o $b$ do item anterior, reescreva $(**)$ na
forma $f''+αf'+βf=0$ $(***)$ e verifique que $f(x) = e^{-x} \cos 2x$
obedece $(***)$.

% (find-angg ".emacs.papers" "hernandez")


\bsk

3) \T(Total: 5.0 pts) Seja $(****)$ esta EDO: $y' = \frac{\cos
  x}{2y}$.

a) \B(1.0 pts) Encontre a solução geral de $(****)$.

b) \B(1.0 pts) Encontre a solução $y=f(x)$ de $(****)$ na qual $f(2π)=-3$.

c) \B(1.0 pts) Encontre uma solução $y=f(x)$ de $(****)$ na qual $f(π)=1$.

d) \B(2.0 pts) Represente graficamente 4 soluções de $(****)$. Dica: calcule o
comportamento delas em $x=k\fracπ2$ e improvise o resto.


\newpage

Gabarito:

(Versão preliminar, não revisado)

\bsk

\def\e#1{e^{-#1x}}

1) $\begin{array}[t]{rcl}
    0 &=& f'' + 5f' + 6f \\
      &=& (D^2 + 5D + 6) f \\
      &=& (D+2) (D=3) f \\
    \end{array}
   $

1a) $f_1(x) = \e2$, $f_2(x) = \e3$

1b) Queremos $f_3(x) = af_1(x) + bf_2(x)$, com

$\begin{array}[t]{rcl}
 f_3(0) = 0 &=& a·1 + b·1 = a+b \quad \funto \quad b=-a \\
 f_3(1) = 1 &=& ae^{-2} + be^{-3} = ae^{-2} - ae^{-3} \\
          a &=& \frac{1}{e^{-2}-e^{-3}} \\
     b = -a &=& \frac{1}{e^{-3}-e^{-2}} \\
 \end{array}
$

1c) Queremos $f_4(x) = cf_1(x) + df_2(x)$, com

$\begin{array}[t]{rcl}
 f_4(0) = 1 &=& c+d \quad \funto \quad d=-1-c \\
 f_4(1) = 0 &=& ce^{-2} + de^{-3} \\
            &=& ce^{-2} + (1-c)e^{-3} \\
            &=& c(e^{-2}-e^{-3}) + e^{-3} \\
          c &=& - \frac{e^{-3}}{e^{-2}-e^{-3}} \\
            &=& \frac{e^{-3}}{e^{-3}-e^{-2}} \\
          d &=& 1 - \frac{e^{-3}}{e^{-3}-e^{-2}} \\
            &=& \frac{(e^{-3}-e^{-2}) - e^{-3}} {e^{-3}-e^{-2}} \\
            &=& \frac{e^{-2}} {e^{-2}-e^{-3}} \\
 \end{array}
$

1d) $f_5(x) = 2f_4(x) + 3f_3(x)$.

\bsk
\bsk

2a) $f_1(x) = e^{(a+ib)x}$, $f_2(x) = e^{(a-ib)x}$, ou

$f_3(x) = \frac{f_1(x)+f_2(x)}{2} = e^{ax} \frac{e^{ibx} + e^{-ibx}} {2} = e^{ax}\cos x$,

$f_4(x) = \frac{f_1(x)-f_2(x)}{2i} = e^{ax} \frac{e^{ibx} - e^{-ibx}} {2i} = e^{ax}\sen x$,

2b) $a=-1$, $b=2$.

$\begin{array}[t]{rcl}
 0 &=& (D-(a+ib)) (D-(a-ib)) f \\
   &=& (D-(-1+2i)) (D-(-1-2i)) f \\
   &=& (D^2 -(-1+2i)D-(-1-2i)D + (-1+2i)(-1-2i)) f \\
   &=& (D^2 +2D + (1+4)) f \\
   &=& f'' + 2f' + 5f \\
 \end{array}
$

$\begin{array}[t]{rcl}
 D(e^{-x} \cos 2x) &=& - (e^{-x} \cos 2x) - 2 (e^{-x} \sen 2x) \\
 D(e^{-x} \sen 2x) &=& - (e^{-x} \sen 2x) + 2 (e^{-x} \cos 2x) \\
 \end{array}
$

Sejam $C = e^{-x} \cos 2x$ e $S = e^{-x} \sen 2x$.

Então $DC = -C-2S$, $DS = -S + 2C = 2C-S$,

$D(DC) = -DC-2DS = -(-C-2S)-2(2C-S) = C+2S-4C+2S = -3C+4S$,

$(D^2 +2D + 5)C = (-3C+4S) +2(-C-2S) + 5C = 0$.






$\begin{array}[t]{l}
     \end{array}
    $






\end{document}



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