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% http://angg.twu.net/LATEX/2016-2-C2-P2.pdf
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% ____ _ _ _
% / ___|__ _| |__ ___ ___ __ _| | |__ ___
% | | / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \
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{\setlength{\parindent}{0em}
\footnotesize
\par Cálculo 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} (página 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
1) \T(Total: 4.0 pts) % Sejam:
a) \B(1.0 pts) Encontre as duas soluções da forma $e^{(a+ib)x}$ de $f''+6f'+13f=0$.
b) \B(0.5 pts) Encontre as duas soluções da forma $e^{(a+ib)x}$ de $f''+3f'-10f=0$.
c) \B(1.0 pts) Encontre $a$ e $b$ para os quais $e^{-x} \sen x$ é solução de de $f''+af'+bf=0$.
d) \B(1.0 pts) Encontre as soluções básicas reais de $f''+6f'+13f=0$.
e) \B(0.5 pts) Encontre uma solução 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 variáveis separáveis.
b) \B(0.5 pts) Teste a sua solução geral.
c) \B(0.5 pts) Encontre uma solução com $f(0)=0$.
\bsk
\bsk
3) \T(Total: 3.0 pts) Prove que a EDO abaixo é 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 é $\ddx \ln(h(x))$? 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 solução pra ela que tenha $f(1)=1$ e
represente graficamente esta solução sobre o campo vetorial do item
anterior.
\newpage
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Mini-gabarito (incompleto e ainda não revisado):
\msk
1a) $0 = f'' + 6f' + 13f
= (D^2 + 6D + 13)f
= (D^2 + (2·3)D + (2^2+3^2)) f$
$= (D + (3+2i)) (D + (3-2i)) f$
$(D + (3+2i))f = 0 \;\; ⇒ \;\; f_1 = e^{-(3+2i)x} = e^{(-3-2i)x}$
$(D + (3-2i))f = 0 \;\; ⇒ \;\; f_2 = e^{-(3-2i)x} = e^{(-3+2i)x}$
\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) $f_1 = e^{-5x}$, $f_2 = e^{2x}$
$f_3(x) = a e^{-5x} + b e^{2x}$
$f'_3(x) = -5a e^{-5x} + 2b e^{2x}$
$f_3(0) = a+b \;\;(=0)$
$f'_3(0) = -5a+2b \;\;(=1)$
$a+b = 0 \;\;⇒\;\; b=-a$
$-5a+2b = 1 \;\;⇒\;\; -5a-2a=1 \;\;⇒\;\; a=-\frac17 \;\;⇒\;\; b=\frac17$
$f_3(x) = -\frac17 e^{-5x} + \frac17 e^{2x}$
\bsk
\def\und#1#2{\underbrace{#1}_{#2}}
2) $\und{\sen(2x)}{f(x)} dx = \und{\sen(5y)}{g(y)} dy$
$∫f(x)dx = ∫g(y)dy \;\;⇒\;\; -\frac12 \cos(2x) = -\frac15 \cos(5x) + C_1$
$\cos(5y) = \frac52 \cos(2x)+C_2$
$5y = \arccos(\frac52 \cos(2x)+C_2)$
$y = \frac15 \arccos(\und{\frac52 \cos(2x)+C_2}{h(x)})$
$\frac{dy}{dx} = \frac15 \arccos'(h(x)) h'(x) = \frac15$
\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 é $z_x dx + z_y dy = 0$, e temos $(z_x)_y \neq (z_y)_x$...
era pra ela ser exata, mas cometi algum erro de digitação...
\bsk
4a) $f'(x) = a$ em $y=2+ax$.
4b) Em $(x,y)=(1,1)$ temos $f'(x)=-1$. A solução é $y=2-x$.
% 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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