Warning: this is an htmlized version!
The original is across this link,
and the conversion rules are here.
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%   ____      _                    _ _           
%  / ___|__ _| |__   ___  ___ __ _| | |__   ___  
% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
%  \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/ 
%                                                

{\setlength{\parindent}{0em}
\footnotesize
\par CÃlculo 2
\par PURO-UFF - 2016.2
\par P2 - 21/dez/2016 - Eduardo Ochs
% \par VersÃo: 14/mar/2016
\par Links importantes:
\par \url{http://angg.twu.net/2016.2-C2.html} (pÃgina do curso)
\par \url{http://angg.twu.net/2016.2-C2/2016.2-C2.pdf} (quadros)
% \par \url{http://angg.twu.net/LATEX/2016-1-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)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar

\bsk
\bsk


2) \T(Total: 4.0 pts) % Sejam:

a) \B(1.0 pts) Encontre as duas soluÃÃes da forma $e^{(a+ib)x}$ de $f''+6f'+13f=0$.

b) \B(0.5 pts) Encontre as duas soluÃÃes da forma $e^{(a+ib)x}$ de $f''+3f'-10f=0$.

c) \B(1.0 pts) Encontre $a$ e $b$ para os quais $e^{-x} \sen x$ Ã soluÃÃo de de $f''+af'+bf=0$.

d) \B(1.0 pts) Encontre as soluÃÃes bÃsicas reais de $f''+6f'+13f=0$.

e) \B(0.5 pts) Encontre uma soluÃÃo de $f''+3f'-10f=0$ que obedece $f(0)=0$, $f'(0)=1$.

\bsk
\bsk

2) \T(Total: 2.0 pts) Considere esta EDO: $$\sen(2x)\,dx = \sen(5y)\,dy$$

a) \B(1.0 pts) Resolva-a por variÃveis separÃveis.

b) \B(0.5 pts) Teste a sua soluÃÃo geral.

c) \B(0.5 pts) Encontre uma soluÃÃo com $f(0)=0$.

\bsk
\bsk

3) \T(Total: 3.0 pts) Prove que a EDO abaixo à exata e resolva-a:
%
$$\left( \frac{ xy^2}{xy^2+3} + \ln(xy^2+3) - y \sen x \right) dx + 
  \left( \frac{2x^2y}{xy^2+3} + \ln(xy^2+3) -   \cos x \right) dy = 0 
$$

Dica: quanto à $\ddx \ln(h(x))$? E $\ddx g(x)\ln(h(x))$?

\bsk
\bsk

4) \T(Total: 1.0 pts) Considere a EDO: $f'(x) = (y-2)/x$.

a) \B(0.5 pts) Represente graficamente o campo vetorial associado a ela.

b) \B(0.5 pts) Encontre uma soluÃÃo pra ela que tenha $f(1)=1$ e
represente graficamente esta soluÃÃo sobre o campo vetorial do item
anterior.





\newpage

%   ____       _                _ _        
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%  \____|\__,_|_.__/ \__,_|_|  |_|\__\___/ 
%                                          
Mini-gabarito (incompleto e ainda nÃo revisado):

\msk

1a) $0 = f'' + 6f' + 13f
       = (D^2 + 6D + 13)f
       = (D^2 + (2Â3)D + (2^2+3^2)) f$

      $= (D + (3+2i)) (D + (3-2i)) f$

    $(D + (3+2i))f = 0 \;\; â \;\; f_1 = e^{-(3+2i)x} = e^{(-3-2i)x}$

    $(D + (3-2i))f = 0 \;\; â \;\; f_2 = e^{-(3-2i)x} = e^{(-3+2i)x}$

\msk

1b) $0 = f'' + 3f' -10f
       = (D^2 + 3D - 10)f
       = (D+5) (D-2) f$

$f_1 = e^{-5x}$

$f_2 = e^{2x}$

\msk

1c)

$f_1 = e^{-x} \cos x
     = e^{-x} (\frac{e^{ix} + e^{-ix}}{2})
     = \frac{1}{2}  e^{(-1+i)x} + \frac{1}{2}  e^{(-1-i)x}$

$f_2 = e^{-x} \sen x
     = e^{-x} (\frac{e^{ix} - e^{-ix}}{2i})
     = \frac{1}{2i} e^{(-1+i)x} - \frac{1}{2i} e^{(-1-i)x}$

$f_3 = e^{(-1+i)x}$

$f_4 = e^{(-1-i)x}$

EDO: $0 = (D-(-1+i)) (D-(-1-i) f
        = (D^2 + 2D + (-1+i)(-1-i)) f$

       $= (D^2 + 2D + 2)f$

$a=2$, $b=2$

\msk

1d) $f_1 = e^{-3x} e^{-2ix}$, $f_2 = e^{-3x} e^{2ix}$

$f_3 = e^{-3x} (\frac{e^{2ix} + e^{-2ix}}{2})  = e^{-3x} \cos 2x$

$f_4 = e^{-3x} (\frac{e^{2ix} - e^{-2ix}}{2i}) = e^{-3x} \sen 2x$

\msk

1e) $f_1 = e^{-5x}$, $f_2 = e^{2x}$

$f_3(x)  =   a e^{-5x} +  b e^{2x}$

$f'_3(x) = -5a e^{-5x} + 2b e^{2x}$

$f_3(0)  = a+b \;\;(=0)$

$f'_3(0) = -5a+2b \;\;(=1)$

$a+b = 0 \;\;â\;\; b=-a$

$-5a+2b = 1 \;\;â\;\; -5a-2a=1 \;\;â\;\; a=-\frac17 \;\;â\;\; b=\frac17$

$f_3(x) = -\frac17 e^{-5x} + \frac17 e^{2x}$

\bsk

\def\und#1#2{\underbrace{#1}_{#2}}

2) $\und{\sen(2x)}{f(x)} dx = \und{\sen(5y)}{g(y)} dy$

$â«f(x)dx = â«g(y)dy \;\;â\;\; -\frac12 \cos(2x) = -\frac15 \cos(5x) + C_1$

$\cos(5y) = \frac52 \cos(2x)+C_2$

$5y = \arccos(\frac52 \cos(2x)+C_2)$

$y = \frac15 \arccos(\und{\frac52 \cos(2x)+C_2}{h(x)})$

$\frac{dy}{dx} = \frac15 \arccos'(h(x)) h'(x) = \frac15$ 



\newpage

3) Sejam:

$z_x = \frac{ xy^2}{xy^2+3} + \ln(xy^2+3) - y \sen x$,

$z_y = \frac{2x^2y}{xy^2+3} + \ln(xy^2+3) -   \cos x$.

A EDO Ã $z_x dx + z_y dy = 0$, e temos $(z_x)_y \neq (z_y)_x$...

era pra ela ser exata, mas cometi algum erro de digitaÃÃo...

\bsk

4a) $f'(x) = a$ em $y=2+ax$.

4b) Em $(x,y)=(1,1)$ temos $f'(x)=-1$. A soluÃÃo à $y=2-x$.



% In [6]: zx = (x*y**2) / (x*y**2+3) + ln(x*y**2+3) - y*sin(x)
% 
% In [7]: zx
% Out[7]: 
%      2                             
%   x*y                    /   2    \
% -------- - y*sin(x) + log\x*y  + 3/
%    2                               
% x*y  + 3                           
% 
% In [8]: zy = (2*x**2*y) / (x*y**2+3) + ln(x*y**2+3) + cos(x)
% 
% In [9]: zy
% Out[9]: 
%     2                            
%  2*x *y       /   2    \         
% -------- + log\x*y  + 3/ + cos(x)
%    2                             
% x*y  + 3                         
% 
% In [10]: zx.diff(y)
% Out[10]: 
%        2  3                      
%     2*x *y       4*x*y           
% - ----------- + -------- - sin(x)
%             2      2             
%   /   2    \    x*y  + 3         
%   \x*y  + 3/                     
% 
% In [11]: zy.diff(x)
% Out[11]: 
%        2  3                    2            
%     2*x *y       4*x*y        y             
% - ----------- + -------- + -------- - sin(x)
%             2      2          2             
%   /   2    \    x*y  + 3   x*y  + 3         
%   \x*y  + 3/                                
% 
% In [12]: zx.diff(y) - zy.diff(x)
% Out[12]: 
%     2   
%   -y    
% --------
%    2    
% x*y  + 3






\end{document}

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