Equações Diferenciais Ordinárias

4 Sistema de EDOs lineares de ordem 1 4 Sistema de EDOs lineares de ordem 1 4.2 Sistema de equações não homogêneas

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4.1 Sistema de equações homogêneas

Nesta seção, discutimos sobre um método de solução para sistemas de EDOs de primeira ordem, lineares, com coeficientes constantes e homogêneas. Ou seja, sistema da forma

{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y}% ^{\prime}(t)=A\boldsymbol{y}(t)},

(4.5)

onde $\boldsymbol{y}(t)=\left(y_{1}(t),y_{2}(t),\dotsc,y_{n}(t)\right)$ , $n>1$ , é o vetor das incógnitas e $A=[a_{ij}]_{i,j=1}^{n,n}$ é a matriz dos coeficientes.

O método consiste em buscar soluções da forma

{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y}% =\boldsymbol{v}e^{rt}},

(4.6)

onde $r$ e o vetor constante $\boldsymbol{v}=(v_{1},v_{2},\cdots,v_{n})$ devem ser determinados.

Substituindo em (4.5), obtemos¹¹1 $\underline{0}=(0,0,\cdots 0)$ .

	$\displaystyle r\boldsymbol{v}e^{rt}=A\boldsymbol{v}e^{rt}$		(4.7)
	$\displaystyle A\boldsymbol{v}e^{rt}-r\boldsymbol{v}e^{rt}=\underline{0}$		(4.8)
	$\displaystyle(A-rI)\boldsymbol{v}\underbrace{e^{rt}}_{>0}=\underline{0},$		(4.9)

ou seja, temos que $r$ e $\boldsymbol{v}$ devem tais que

{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}(A-rI)% \boldsymbol{v}=\underline{0}}.

(4.10)

Em outras palavras, $r$ é autovalor e $\boldsymbol{v}$ é autovetor da matriz $A$ .

Com isso, concluímos que se $r_{1}$ autovalor e $\boldsymbol{v}_{1}$ autovetor de $A$ , então

\boldsymbol{y}_{1}(t)=\boldsymbol{v}_{1}e^{r_{1}t}

(4.11)

é solução particular de (4.5). A solução geral tem a forma

{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y}% (t)=c_{1}\boldsymbol{y_{1}}(t)+c_{2}\boldsymbol{y_{2}}(t)+\cdots+c_{n}% \boldsymbol{y_{n}}(t)},

(4.12)

onde $\boldsymbol{y_{1}}$ , $\boldsymbol{y_{2}}$ , $\dotsc$ , $\boldsymbol{y}_{n}$ formam um conjunto fundamental de soluções de (4.5), i.e. são soluções linearmente independentes.

4.1.1 Autovalores reais distintos

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No caso da matriz dos coeficientes $A$ ter apenas todos os autovalores reais e dois a dois distintos, então a solução geral de (4.5) é

{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y}% (t)=c_{1}\boldsymbol{v}_{1}e^{r_{1}t}+c_{2}\boldsymbol{v}_{2}e^{r_{2}t}+\cdots% +c_{n}\boldsymbol{v}_{n}e^{r_{n}t}},

(4.13)

onde $r_{i}$ e $\boldsymbol{v}_{i}$ é autovalor e autovetor de $A$ respectivamente e $i=1,2,\cdots,n$ . A independência linear das soluções é garantida pelo wronskiano

W(e^{r_{1}t},\dotsc,e^{r_{n}t};t)\neq 0.

(4.14)

Exemplo 4.1.1.

Vamos resolver o seguinte sistema

	$\displaystyle y_{1}^{\prime}(t)$	$\displaystyle=-2y_{1}(t)+2y_{2}(t)$		(4.15)
	$\displaystyle y_{2}^{\prime}(t)$	$\displaystyle=-2y_{1}(t)+3y_{2}(t).$		(4.16)

Primeiramente, reescrevemos o sistema na sua forma matricial

\boldsymbol{y}^{\prime}=\underbrace{\begin{bmatrix}-2&2\\ -2&3\end{bmatrix}}_{A}\boldsymbol{y},

(4.17)

onde $\boldsymbol{y}:t\mapsto\boldsymbol{y}(t)=(y_{1}(t),y_{2}(t))$ . Então, calculamos os autovalores da matriz dos coeficientes $A$ . Para tanto, resolvemos sua equação característica

	$\displaystyle\|A-rI\|=0$		(4.18)
	$\displaystyle\left\|\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-2&2\\ \hidden@noalign{}\hfil\textstyle-2&3\end{bmatrix}-r\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1&0\\ \hidden@noalign{}\hfil\textstyle 0&1\end{bmatrix}\right\|=0$		(4.23)
	$\displaystyle\begin{vmatrix}\hidden@noalign{}\hfil\textstyle-2-r&2\\ \hidden@noalign{}\hfil\textstyle-2&3-r\end{vmatrix}=0$		(4.26)
	$\displaystyle(-2-r)(3-r)+4=0$		(4.27)
	$\displaystyle r^{2}-r-2=0,$		(4.28)
	$\displaystyle r=\frac{1\pm\sqrt{9}}{2}$		(4.29)
	$\displaystyle r_{1}=-1,\quad r_{2}=2.$		(4.30)

Obtidos os autovalores, calculamos os autovetores $\boldsymbol{v}_{1}$ e $\boldsymbol{v}_{2}$ . Começamos calculando o autovetor associado a $r_{1}$ .

	$\displaystyle(A-r_{1}I)\boldsymbol{v}_{1}=\underline{0}$		(4.31)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-2-r_{1}&2\\ \hidden@noalign{}\hfil\textstyle-2&3-r_{1}\end{bmatrix}\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle v_{11}\\ \hidden@noalign{}\hfil\textstyle v_{21}\end{bmatrix}=\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}$		(4.38)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1&2\\ \hidden@noalign{}\hfil\textstyle-2&4\end{bmatrix}\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle v_{11}\\ \hidden@noalign{}\hfil\textstyle v_{21}\end{bmatrix}=\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}$		(4.45)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1&2&\|&0\\ \hidden@noalign{}\hfil\textstyle-2&4&\|&0\end{bmatrix}\begin{matrix}% \hidden@noalign{}\hfil\textstyle\times 2\\ \hidden@noalign{}\hfil\textstyle\leftarrow-\end{matrix}$		(4.50)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1&2&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&\|&0\end{bmatrix}$		(4.53)

Assim, temos

	$\displaystyle-v_{11}+2v_{21}=0$		(4.54)
	$\displaystyle v_{11}=2v_{21},$		(4.55)

donde escolhemos

\boldsymbol{v}_{1}=\begin{bmatrix}2\\ 1\end{bmatrix}.

(4.56)

Com isso, temos obtido a solução particular

	$\displaystyle\boldsymbol{y}_{1}(t)$	$\displaystyle=\boldsymbol{v}_{1}e^{r_{1}t}$		(4.57)
		$\displaystyle=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}e^{-t}$		(4.60)

Agora, de forma análoga, calculamos $\boldsymbol{v}_{2}$ , um autovetor associado a $r_{2}$ .

	$\displaystyle(A-r_{2}I)\boldsymbol{v}_{2}=\underline{0}$		(4.61)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-2-r_{2}&2&\|&0\\ \hidden@noalign{}\hfil\textstyle-2&3-r_{2}&\|&0\end{bmatrix}$		(4.64)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-4&2&\|&0\\ \hidden@noalign{}\hfil\textstyle-2&1&\|&0\end{bmatrix}\begin{matrix}% \hidden@noalign{}\hfil\textstyle\leftarrow-\\ \hidden@noalign{}\hfil\textstyle\times 2\end{matrix}$		(4.69)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&0&\|&0\\ \hidden@noalign{}\hfil\textstyle-2&1&\|&0\end{bmatrix}$		(4.72)
	$\displaystyle-2v_{12}+v_{22}=0$		(4.73)
	$\displaystyle\boldsymbol{v}_{2}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle 2\end{bmatrix}$		(4.76)

Com isso, temos a solução particular

	$\displaystyle\boldsymbol{y}_{2}(t)$	$\displaystyle=\boldsymbol{v}_{2}e^{r_{2}t}$		(4.77)
		$\displaystyle=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 2\end{bmatrix}e^{2t}$		(4.80)

De tudo isso, concluímos que a solução geral de (4.15) é

	$\displaystyle\boldsymbol{y}(t)$	$\displaystyle=c_{1}\boldsymbol{y}_{1}(t)+c_{2}\boldsymbol{y}_{2}$		(4.81)
		$\displaystyle=c_{1}\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}e^{-t}+c_{2}\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 2\end{bmatrix}e^{2t}$		(4.86)

ou, ainda,

	$\displaystyle y_{1}(t)$	$\displaystyle=2c_{1}e^{-t}+c_{2}e^{2t}$		(4.87)
	$\displaystyle y_{2}(t)$	$\displaystyle=c_{1}e^{-t}+2c_{2}e^{2t}$		(4.88)

No Python, podemos computar os autovalores e autovetores da matriz $A$ com os seguintes comandos²²2Veja mais informações em SymPy: Matrices.:

⬇

1 In : from sympy import *

2 In : A = Matrix([[-2,2],[-2,3]])

3 In : A.eigenvects()

4 Out:

5 [(-1, 1, [Matrix([

6 [2],

7 [1]])]), (2, 1, [Matrix([

8 [1/2],

9 [ 1]])])]

Então, a solução do sistema (4.15) pode ser computada com os comandos:

⬇

1 In : t,C1,C2 = symbols('t,C1,C2')

2 In : y1,y2 = symbols('y1,y2', cls=Function)

3 In : se = (Eq(diff(y1(t),t),-2*y1(t)+2*y2(t)), \

4 ...: Eq(diff(y2(t),t),-2*y1(t)+3*y2(t)))

5 ...:

6 In : dsolve(se,[y1(t),y2(t)])

7 Out:

8 [Eq(y1(t), 2*C1*exp(-t) + 2*C2*exp(2*t)),

9 Eq(y2(t), C1*exp(-t) + 4*C2*exp(2*t))]

4.1.2 Autovalores reais repetidos

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Um autovalor real duplo $r$ da matriz de coeficientes $A$ nos fornece duas soluções particulares para (4.5), a saber

	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y% }_{1}(t)}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=\boldsymbol{% v}_{1}e^{rt}},$		(4.89)
	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y% }_{2}(t)}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=\boldsymbol{% v}_{1}te^{rt}+\boldsymbol{v}_{2}e^{rt}},$		(4.90)

onde $\boldsymbol{v}_{1}$ é autovetor associado a $r$ . Para encontrar $\boldsymbol{v}_{2}$ , substituímos $\boldsymbol{y}_{2}$ em (4.5), donde

	$\displaystyle\boldsymbol{y}_{2}^{\prime}=A\boldsymbol{y}_{2}$		(4.91)
	$\displaystyle\boldsymbol{v}_{1}e^{rt}+r\boldsymbol{v}_{1}te^{rt}+r\boldsymbol{% v}_{2}e^{rt}=A\boldsymbol{v}_{1}te^{rt}+A\boldsymbol{v}_{2}e^{rt}$		(4.92)
	$\displaystyle(A-rI)\boldsymbol{v}_{1}t+(A-rI)\boldsymbol{v}_{2}=\boldsymbol{v}% _{1}$		(4.93)

Segue que

	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}(A-rI)% \boldsymbol{v}_{1}}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=\underline{0}}$		(4.94)
	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}(A-rI)% \boldsymbol{v}_{2}}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=\boldsymbol{% v}_{1}}$		(4.95)

Exemplo 4.1.2.

Vamos calcular a solução geral de

\boldsymbol{y}^{\prime}=\begin{bmatrix}-1&1&0\\ 0&-1&-1\\ 0&0&1\end{bmatrix}\boldsymbol{y}.

(4.96)

Neste caso, temos que $r_{1,2}=-1$ é autovalor duplo e $r_{3}=1$ é autovalor simples da matriz de coeficientes do sistema (4.96).

Associadas a $r_{1,2}$ buscamos por soluções particulares da forma

	$\displaystyle\boldsymbol{y}_{1}=\boldsymbol{v}_{1}e^{r_{1,2}t},$		(4.97)
	$\displaystyle\boldsymbol{y}_{2}=\boldsymbol{v}_{1}te^{r_{1,2}t}+\boldsymbol{v}% _{2}e^{r_{1,2}t}.$		(4.98)

Calculamos $\boldsymbol{v}_{1}$ resolvendo

	$\displaystyle(A-r_{1,2}I)\boldsymbol{v}_{1}=\underline{0}$		(4.99)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1-r_{1,2}&1&0&\|&0% \\ \hidden@noalign{}\hfil\textstyle 0&-1-r_{1,2}&-1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&1-r_{1,2}&\|&0\end{bmatrix}$		(4.103)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&1&0&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&-1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&2&\|&0\end{bmatrix}$		(4.107)

Com isso, podemos escolher

\boldsymbol{v}_{1}=\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}.

(4.108)

Determinado $\boldsymbol{v}_{1}$ , calculamos $\boldsymbol{v}_{2}$ com

	$\displaystyle(A-r_{1,2}I)\boldsymbol{v}_{2}=\boldsymbol{v}_{1}$		(4.109)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&1&0&\|&1\\ \hidden@noalign{}\hfil\textstyle 0&0&-1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&2&\|&0\end{bmatrix}$		(4.113)

donde escolhemos

\boldsymbol{v}_{2}=\begin{bmatrix}1\\ 1\\ 0\end{bmatrix}.

(4.114)

Agora, associada a $r_{3}$ temos uma solução particular da forma

\boldsymbol{y}_{3}=\boldsymbol{v}_{3}e^{r_{3}t},

(4.115)

onde

	$\displaystyle(A-r_{3}I)\boldsymbol{v}_{3}=\underline{0}$		(4.116)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1-r_{3}&1&0&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&-1-r_{3}&-1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&1-r_{3}&\|&0\end{bmatrix}$		(4.120)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-2&1&0&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&-2&-1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&\|&0\end{bmatrix}$		(4.124)

Com isso, temos $-2v_{13}+v_{23}=0$ , $-2v_{23}-v_{33}=0$ . Ou seja, podemos escolher

\boldsymbol{v_{3}}=\begin{bmatrix}-1\\ -2\\ 4\end{bmatrix}.

(4.125)

Com tudo isso, podemos concluir que a solução geral de (4.96) é

	$\displaystyle\boldsymbol{y}(t)$	$\displaystyle=c_{1}\boldsymbol{v}_{1}e^{r_{1,2}t}+c_{2}\left(\boldsymbol{v}_{1% }t+\boldsymbol{v}_{2}\right)e^{r_{1,2}t}+c_{3}\boldsymbol{v}_{3}e^{r_{3}t}$		(4.126)
	$\displaystyle\boldsymbol{y}(t)$	$\displaystyle=c_{1}\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}e^{-t}+c_{2}\left(\begin{% bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}t+\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\right)e^{-t}+c_{3}\begin{% bmatrix}\hidden@noalign{}\hfil\textstyle-1\\ \hidden@noalign{}\hfil\textstyle-2\\ \hidden@noalign{}\hfil\textstyle 4\end{bmatrix}e^{t}.$		(4.139)

4.1.3 Autovalores complexos

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Vejamos o caso de ${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}r=\lambda\pm% \mu i}$ serem autovalores complexos da matriz de coeficientes do sistema (4.5). Associados, temos autovetores da forma

	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{v}}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=\boldsymbol{% v}_{1}+\boldsymbol{v}_{2}i},$		(4.140)
	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\overline{% \boldsymbol{v}}}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=\boldsymbol{% v}_{1}-\boldsymbol{v}_{2}i}.$		(4.141)

Por substituição direta, podemos verificar que

	$\displaystyle\boldsymbol{u}_{1}(t)$	$\displaystyle=\boldsymbol{v}e^{(\lambda+\mu i)t},$		(4.142)
	$\displaystyle\boldsymbol{u}_{2}(t)$	$\displaystyle=\overline{\boldsymbol{v}}e^{(\lambda-\mu i)t},$		(4.143)

são soluções de (4.5). Também, verifica-se que as partes real e imaginária de $\boldsymbol{u}_{1}$ e ${u}_{2}$ são soluções reais de (4.5). A fim determiná-las, usamos a fórmula de Euler³³3Leonhard Paul Euler, 1707-1783, matemático e físico suíço. Fonte: Wikipédia: Leonhard Euler., calculamos

$\displaystyle\boldsymbol{u}_{1}$	$\displaystyle=(\boldsymbol{v}_{1}+\boldsymbol{v}_{2}i)e^{(\lambda+\mu i)t}$	(4.144)
$\displaystyle\boldsymbol{u}_{1}$	$\displaystyle=(\boldsymbol{v}_{1}+\boldsymbol{v}_{2}i)e^{\lambda t}[\cos(\mu t% )+i\operatorname{sen}(\mu t)]$	(4.145)
$\displaystyle\boldsymbol{u}_{1}$	$\displaystyle=e^{\lambda t}[\boldsymbol{v_{1}}\cos(\mu t)-\boldsymbol{v}_{2}% \operatorname{sen}(\mu t)]$	(4.146)
	$\displaystyle=ie^{\lambda t}[\boldsymbol{v_{1}}\operatorname{sen}(\mu t)+% \boldsymbol{v}_{2}\cos(\mu t)].$	(4.147)

De forma análoga, verifica-se que $\boldsymbol{u}_{1}=\overline{\boldsymbol{u}_{2}}$ . Ou seja, as partes reais e imaginárias de $\boldsymbol{u}_{1}$ e $\boldsymbol{u}_{2}$ são

	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y% }_{1}(t)}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=e^{\lambda t% }[\boldsymbol{v_{1}}\cos(\mu t)-\boldsymbol{v}_{2}\operatorname{sen}(\mu t)]},$		(4.148)
	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}\boldsymbol{y% }_{2}(t)}$	$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1% }\pgfsys@color@rgb@stroke{0}{0}{1}\pgfsys@color@rgb@fill{0}{0}{1}=e^{\lambda t% }[\boldsymbol{v_{1}}\operatorname{sen}(\mu t)+\boldsymbol{v}_{2}\cos(\mu t)]},$		(4.149)

as quais são soluções linearmente independentes, particulares de (4.5).

Exemplo 4.1.3.

Vamos calcular a solução geral de

\boldsymbol{y}^{\prime}=\begin{bmatrix}1&1\\ -1&1\end{bmatrix}\boldsymbol{y}.

(4.150)

Começamos calculando os autovalores associados a matriz de coeficientes do sistema (4.150). Podemos fazer isso como segue

	$\displaystyle\|A-rI\|=0$		(4.151)
	$\displaystyle\begin{vmatrix}\hidden@noalign{}\hfil\textstyle 1-r&1\\ \hidden@noalign{}\hfil\textstyle-1&1-r\end{vmatrix}=0$		(4.154)
	$\displaystyle(1-r)^{2}+1=0$		(4.155)
	$\displaystyle r^{2}-2r+2=0$		(4.156)
	$\displaystyle r=1\pm i$		(4.157)

Um autovetor associado a $r=1+i$ pode ser obtidos resolvendo-se

	$\displaystyle(A-rI)\boldsymbol{v}=\underline{0}\begin{bmatrix}\hidden@noalign{% }\hfil\textstyle 1-r&1&\|&0\\ \hidden@noalign{}\hfil\textstyle-1&1-r&\|&0\end{bmatrix}$		(4.160)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-i&1&\|&0\\ \hidden@noalign{}\hfil\textstyle-1&-i&\|&0\end{bmatrix}\begin{matrix}% \hidden@noalign{}\hfil\textstyle\times i\\ \hidden@noalign{}\hfil\textstyle\leftarrow+\end{matrix}$		(4.165)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-i&1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&\|&0\end{bmatrix}$		(4.168)
	$\displaystyle-iv_{1}+v_{2}=0.$		(4.169)

Com isso, podemos escolher o autovetor

	$\displaystyle\boldsymbol{v}$	$\displaystyle=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle i\end{bmatrix}$		(4.172)
		$\displaystyle=\underbrace{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}}_{\boldsymbol{v}_{1}}+i% \underbrace{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}}_{\boldsymbol{v}_{2}}$		(4.177)

Desta forma, identificamos as soluções particulares

	$\displaystyle\boldsymbol{y}_{1}(t)$	$\displaystyle=e^{t}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\cos(t)-\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}\operatorname{sen}(t)\right\}$		(4.182)
	$\displaystyle\boldsymbol{y}_{2}(t)$	$\displaystyle=e^{t}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\operatorname{sen}(t)+\begin{% bmatrix}\hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}\cos(t)\right\}$		(4.187)

Concluímos que a solução geral de (4.150) é

$\displaystyle\boldsymbol{y}(t)$	$\displaystyle=c_{1}\boldsymbol{y}_{1}(t)+c_{2}\boldsymbol{y}_{2}(t)$	(4.188)
	$\displaystyle=c_{1}e^{t}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1% \\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\cos(t)-\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}\operatorname{sen}(t)\right\}$	(4.193)
	$\displaystyle+c_{2}e^{t}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1% \\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\operatorname{sen}(t)+\begin{% bmatrix}\hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}\cos(t)\right\}$	(4.198)

Exercícios resolvidos

ER 4.1.1.

Resolva o seguinte PVI

	$\displaystyle\boldsymbol{y}^{\prime}$	$\displaystyle=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-3&1\\ \hidden@noalign{}\hfil\textstyle 1&-3\end{bmatrix}\boldsymbol{y},$		(4.201)
	$\displaystyle\boldsymbol{y}(0)$	$\displaystyle=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}$		(4.204)

Resolução.

O primeiro passo é encontrar a solução geral de

\boldsymbol{y}^{\prime}=\begin{bmatrix}-3&1\\ 1&-3\end{bmatrix}\boldsymbol{y}.

(4.205)

Para tanto, calculamos os autovalores da matriz dos coeficientes.

	$\displaystyle\|A-rI\|=0$		(4.206)
	$\displaystyle\begin{vmatrix}\hidden@noalign{}\hfil\textstyle-3-r&1\\ \hidden@noalign{}\hfil\textstyle 1&-3-r\end{vmatrix}=0$		(4.209)
	$\displaystyle(-3-r)^{2}-1=0$		(4.210)
	$\displaystyle r^{2}+6r+8=0$		(4.211)
	$\displaystyle r_{1}=-4,\quad r_{2}=-2.$		(4.212)

Então, buscamos por autovetores associados.

	$\displaystyle(A-r_{1}I)\boldsymbol{v}_{1}=\underline{0}$		(4.213)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1&1&\|&0\\ \hidden@noalign{}\hfil\textstyle 1&1&\|&0\end{bmatrix}$		(4.216)
	$\displaystyle v_{11}+v_{21}=0$		(4.217)
	$\displaystyle\boldsymbol{v}_{1}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle-1\end{bmatrix}$		(4.220)

	$\displaystyle(A-r_{2}I)\boldsymbol{v}_{2}=\underline{0}$		(4.221)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1&1&\|&0\\ \hidden@noalign{}\hfil\textstyle 1&-1&\|&0\end{bmatrix}\begin{matrix}% \hidden@noalign{}\hfil\textstyle\times 1\\ \hidden@noalign{}\hfil\textstyle\leftarrow+\end{matrix}$		(4.226)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-1&1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&\|&0\end{bmatrix}-v_{12}+v_{22}=0$		(4.229)
	$\displaystyle\boldsymbol{v}_{2}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}$		(4.232)

Com isso, temos a solução geral da EDO dada por

	$\displaystyle\boldsymbol{y}(t)$	$\displaystyle=c_{1}\boldsymbol{v}_{1}e^{r_{1}t}+c_{2}\boldsymbol{v}_{2}e^{r_{2% }t}$		(4.233)
		$\displaystyle=c_{1}\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle-1\end{bmatrix}e^{-4t}+c_{2}\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}e^{-2t}.$		(4.238)

Agora, aplicamos a condição inicial.

	$\displaystyle\boldsymbol{y}(0)=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle% -1\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}$		(4.241)
	$\displaystyle c_{1}\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle-1\end{bmatrix}+c_{2}\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}=\begin{bmatrix}\hidden@noalign% {}\hfil\textstyle-1\\ \hidden@noalign{}\hfil\textstyle 1\end{bmatrix}$		(4.248)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1&1&\|&-1\\ \hidden@noalign{}\hfil\textstyle-1&1&\|&1\end{bmatrix}\begin{matrix}% \hidden@noalign{}\hfil\textstyle\times 1\\ \hidden@noalign{}\hfil\textstyle\leftarrow+\end{matrix}$		(4.253)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1&1&\|&-1\\ \hidden@noalign{}\hfil\textstyle 0&2&\|&0\end{bmatrix}$		(4.256)

Segue que

	$\displaystyle c_{1}+c_{2}$	$\displaystyle=-1$		(4.257)
	$\displaystyle 2c_{2}$	$\displaystyle=0$		(4.258)

donde, $c_{1}=-1$ e $c_{2}=0$ .

Concluímos que a solução do PVI é

\boldsymbol{y}(t)=\begin{bmatrix}-1\\ 1\end{bmatrix}e^{-4t}.

(4.259)

No Python, podemos computar a solução deste exercício com os seguintes comandos:

⬇

1 In : from sympy import *

2 In : t,C1,C2 = symbols('t,C1,C2')

3 In : y1,y2 = symbols('y1,y2', cls=Function)

4 In : ed = (Eq(diff(y1(t),t),-3*y1(t)+y2(t)),\

5 ...: Eq(diff(y2(t),t),y1(t)-3*y2(t)))

6 ...:

7 In : ds = dsolve(ed,[y1(t),y2(t)])

8 In : cs = solve((Eq(ds[0].rhs.subs(t,0),-1), \

9 ...: Eq(ds[1].rhs.subs(t,0),1)),[C1,C2])

10 ...:

11 In : ds[0].subs(cs)

12 Out: Eq(y1(t), -exp(-4*t))

14 In : ds[1].subs(cs)

15 Out: Eq(y2(t), exp(-4*t))

ER 4.1.2.

Calcule a solução geral do seguinte sistema

$\displaystyle y_{1}^{\prime}$	$\displaystyle=-y_{1}+2y_{2}$	(4.260)
$\displaystyle y_{2}^{\prime}$	$\displaystyle=-2y_{1}-y_{2}+y_{3}$	(4.261)
$\displaystyle y_{3}^{\prime}$	$\displaystyle=-y_{3}+y_{4}$	(4.262)
$\displaystyle y_{4}^{\prime}$	$\displaystyle=-y_{4}$	(4.263)

Resolução.

Vamos reescrever o sistema na sua forma matricial.

\boldsymbol{y}^{\prime}=\begin{bmatrix}-1&2&0&0\\ -2&-1&1&0\\ 0&0&-1&1\\ 0&0&0&-1\end{bmatrix}\boldsymbol{y}

(4.264)

Calculamos os autovalores da matriz dos coeficientes.

	$\displaystyle\|A-rI\|=0$		(4.265)
	$\displaystyle\begin{vmatrix}\hidden@noalign{}\hfil\textstyle-1-r&2&0&0\\ \hidden@noalign{}\hfil\textstyle-2&-1-r&1&0\\ \hidden@noalign{}\hfil\textstyle 0&0&-1-r&1\\ \hidden@noalign{}\hfil\textstyle 0&0&0&-1-r\end{vmatrix}=0$		(4.270)
	$\displaystyle r^{4}+4r^{3}+10r^{2}+12r+5=0$		(4.271)

Resolvendo esta equação característica, obtemos os autovalores $r_{1,2}=-1$ e $r_{3},r_{4}=-1\pm 2i$ .

a)

$r_{1},r_{2}=-1$ :

Soluções particulares associadas.

$\displaystyle\boldsymbol{y}_{1}=\boldsymbol{v}_{1}e^{r_{1,2}t},$ (4.272)

$\displaystyle\boldsymbol{y}_{2}=\boldsymbol{v}_{1}te^{r_{1,2}t}+\boldsymbol{v}% _{2}e^{r_{1,2}t}$ (4.273)

O vetor $\boldsymbol{v}_{1}$ é autovetor associado a $r_{1,2}$ .

$\displaystyle(A-r_{1}I)\boldsymbol{v}_{1}=\underline{0}$ (4.274)

$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&2&0&0&|&0\\ \hidden@noalign{}\hfil\textstyle-2&0&1&0&|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&1&|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&0&|&0\end{bmatrix}$ (4.279)

$\displaystyle\boldsymbol{v}_{1}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}$ (4.284)

O vetor $\boldsymbol{v}_{2}$ é calculado como segue.

$\displaystyle(A-r_{1,2}I)\boldsymbol{v}_{2}=\boldsymbol{v}_{1}$ (4.285)

$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&2&0&0&|&1\\ \hidden@noalign{}\hfil\textstyle-2&0&1&0&|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&1&|&2\\ \hidden@noalign{}\hfil\textstyle 0&0&0&0&|&0\end{bmatrix}$ (4.290)

$\displaystyle\boldsymbol{v}_{2}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle\frac{1}{2}\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 2\end{bmatrix}$ (4.295)
b)

$r_{3},r_{4}=-1+2i$ :

Soluções particulares associadas.

$\displaystyle\boldsymbol{y}_{3}$ $\displaystyle=e^{-t}[\boldsymbol{v_{3}}\cos(2t)-\boldsymbol{v}_{4}% \operatorname{sen}(2t)]$ (4.296)

$\displaystyle\boldsymbol{y}_{4}$ $\displaystyle=e^{-t}[\boldsymbol{v_{3}}\operatorname{sen}(2t)+\boldsymbol{v}_{% 4}\operatorname{sen}(2t)]$ (4.297)

Os vetores $\boldsymbol{v}_{3}$ e $\boldsymbol{v}_{4}$ são, respectivamente, as partes real e imaginária de autovetor associado a $r_{3}$ ou $r_{4}$ . Usando $r_{3}$ e denotando o autovetor por $\boldsymbol{v}$ , calculamos como segue.

$\displaystyle(A-r_{3}I)\boldsymbol{v}=\underline{0}$ (4.298)

$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-2i&2&0&0&|&0\\ \hidden@noalign{}\hfil\textstyle-2&-2i&1&0&|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&-2i&1&|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&-2i&|&0\end{bmatrix}$ (4.303)

$\displaystyle\boldsymbol{v}=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1% \\ \hidden@noalign{}\hfil\textstyle i\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}=\underbrace{\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}}_{\boldsymbol{v_{1}}}+i% \underbrace{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}}_{\boldsymbol{v}_{2}}$ (4.316)

De tudo isso, temos a solução geral

$\displaystyle\boldsymbol{y}(t)$	$\displaystyle=c_{1}\boldsymbol{y}_{1}(t)+c_{2}\boldsymbol{y}_{2}+c_{3}% \boldsymbol{y}_{3}(t)+c_{4}\boldsymbol{y}_{4}$	(4.317)
	$\displaystyle=c_{1}\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}e^{-t}$	(4.322)
	$\displaystyle+c_{2}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}t+\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle\frac{1}{2}\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 2\end{bmatrix}\right\}e^{-t}$	(4.331)
	$\displaystyle+c_{3}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\cos(2t)-\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\operatorname{sen}(2t)\right\}$	(4.340)
	$\displaystyle+c_{4}\left\{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\operatorname{sen}(2t)+\begin{% bmatrix}\hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}\cos(2t)\right\}$	(4.349)

ou, equivalentemente,

$\displaystyle y_{1}(t)$	$\displaystyle=c_{1}e^{-t}+c_{2}(t+1)e^{-t}$	(4.350)
	$\displaystyle+c_{3}\cos(2t)+c_{4}\operatorname{sen}(2t)$	(4.351)
$\displaystyle y_{2}(t)$	$\displaystyle=\frac{c_{2}}{2}e^{-t}-c_{3}\operatorname{sen}(2t)+c_{4}\cos(2t)$	(4.352)
$\displaystyle y_{3}(t)$	$\displaystyle=2c_{1}e^{-t}+c_{2}(2t+2)e^{-t}$	(4.353)
$\displaystyle y_{4}(t)$	$\displaystyle=2c_{2}e^{-t}$	(4.354)

Exercícios

E. 4.1.1.

Calcule a solução geral de

	$\displaystyle y_{1}^{\prime}(t)$	$\displaystyle=-y_{1}(t)+2y_{2}(t)$		(4.355)
	$\displaystyle y_{2}^{\prime}(t)$	$\displaystyle=4y_{1}(t)+y_{2}(t)$		(4.356)

$y_{1}(t)=c_{1}e^{-3t}+c_{2}e^{3t}$ , $y_{2}(t)=-c_{1}e^{-3t}+2c_{2}e^{3t}$

E. 4.1.2.

Calcule a solução do PVI

	$\displaystyle y_{1}^{\prime}(t)$	$\displaystyle=2y_{1}(t)+y_{2}(t),\quad y_{1}(0)=0$		(4.357)
	$\displaystyle y_{2}^{\prime}(t)$	$\displaystyle=4y_{1}(t)-y_{2}(t),\quad y_{2}(0)=5$		(4.358)

$y_{1}(t)=e^{3t}-e^{-2t}$ , $y_{2}(t)=e^{3t}+4e^{-2t}$

E. 4.1.3.

Calcule a solução geral de

\boldsymbol{y}^{\prime}=\begin{bmatrix}1&1\\ -1&3\end{bmatrix}\boldsymbol{y}

(4.359)

$\displaystyle\boldsymbol{y}(t)=c_{1}\begin{bmatrix}1\\ 1\end{bmatrix}e^{2t}+c_{2}\begin{bmatrix}1\\ 1\end{bmatrix}te^{2t}+c_{2}\begin{bmatrix}-1\\ 0\end{bmatrix}e^{2t}$

E. 4.1.4.

Calcule a solução do PVI

	$\displaystyle y_{1}^{\prime}(t)$	$\displaystyle=-3y_{1}(t)+y_{2}(t),\quad y_{1}(0)=2$		(4.360)
	$\displaystyle y_{2}^{\prime}(t)$	$\displaystyle=-y_{1}(t)-y_{2}(t),\quad y_{2}(0)=1$		(4.361)

$y_{1}(t)=(-t+2)e^{-2t}$ , $y_{2}(t)=(-t+1)e^{-2t}$

E. 4.1.5.

Encontre a solução geral de

	$\displaystyle y_{1}^{\prime}$	$\displaystyle=2y_{1}-y_{2}$		(4.362)
	$\displaystyle y_{2}^{\prime}$	$\displaystyle=5y_{1}-2y_{2}$		(4.363)

$y_{1}(t)=-c_{1}\operatorname{sen}(t)-c_{2}\cos(t)$ , $y_{2}(t)=(-2c_{1}-c_{2})\operatorname{sen}(t)+(c_{1}-2c_{2})\cos(t)$

E. 4.1.6.

Encontre a solução geral de

$\displaystyle y_{1}^{\prime}$	$\displaystyle=y_{1}+4y_{2}-6y_{3}$	(4.364)
$\displaystyle y_{2}^{\prime}$	$\displaystyle=-y_{1}-3y_{2}+3y_{3}$	(4.365)
$\displaystyle y_{3}^{\prime}$	$\displaystyle=-y_{1}-2y_{2}+2y_{3}$	(4.366)

$y_{1}(t)=-2c_{1}e^{-t}-2c_{2}te^{-t}+3c_{2}e^{-t}-2c_{3}e^{2t}$ , $y_{2}(t)=c_{1}e^{-t}+c_{2}te^{-t}+c_{3}e^{2t}$ , $y_{3}(t)=c_{2}e^{-t}+c_{3}e^{2t}$

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Política de uso de dados

Política de uso de dados

Equações Diferenciais Ordinárias

4.1 Sistema de equações homogêneas

4.1.1 Autovalores reais distintos

Exemplo 4.1.1.

4.1.2 Autovalores reais repetidos

Exemplo 4.1.2.

4.1.3 Autovalores complexos

Exemplo 4.1.3.

Exercícios resolvidos

ER 4.1.1.

Resolução.

ER 4.1.2.

Resolução.

Exercícios

E. 4.1.1.

E. 4.1.2.

E. 4.1.3.

E. 4.1.4.

E. 4.1.5.

E. 4.1.6.

Envie seu comentário

	$\displaystyle\boldsymbol{y}_{1}=\boldsymbol{v}_{1}e^{r_{1,2}t},$		(4.272)
	$\displaystyle\boldsymbol{y}_{2}=\boldsymbol{v}_{1}te^{r_{1,2}t}+\boldsymbol{v}% _{2}e^{r_{1,2}t}$		(4.273)

	$\displaystyle(A-r_{1}I)\boldsymbol{v}_{1}=\underline{0}$		(4.274)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&2&0&0&\|&0\\ \hidden@noalign{}\hfil\textstyle-2&0&1&0&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&0&\|&0\end{bmatrix}$		(4.279)
	$\displaystyle\boldsymbol{v}_{1}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}$		(4.284)

	$\displaystyle(A-r_{1,2}I)\boldsymbol{v}_{2}=\boldsymbol{v}_{1}$		(4.285)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0&2&0&0&\|&1\\ \hidden@noalign{}\hfil\textstyle-2&0&1&0&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&1&\|&2\\ \hidden@noalign{}\hfil\textstyle 0&0&0&0&\|&0\end{bmatrix}$		(4.290)
	$\displaystyle\boldsymbol{v}_{2}=\begin{bmatrix}\hidden@noalign{}\hfil% \textstyle 1\\ \hidden@noalign{}\hfil\textstyle\frac{1}{2}\\ \hidden@noalign{}\hfil\textstyle 2\\ \hidden@noalign{}\hfil\textstyle 2\end{bmatrix}$		(4.295)

	$\displaystyle\boldsymbol{y}_{3}$	$\displaystyle=e^{-t}[\boldsymbol{v_{3}}\cos(2t)-\boldsymbol{v}_{4}% \operatorname{sen}(2t)]$		(4.296)
	$\displaystyle\boldsymbol{y}_{4}$	$\displaystyle=e^{-t}[\boldsymbol{v_{3}}\operatorname{sen}(2t)+\boldsymbol{v}_{% 4}\operatorname{sen}(2t)]$		(4.297)

	$\displaystyle(A-r_{3}I)\boldsymbol{v}=\underline{0}$		(4.298)
	$\displaystyle\begin{bmatrix}\hidden@noalign{}\hfil\textstyle-2i&2&0&0&\|&0\\ \hidden@noalign{}\hfil\textstyle-2&-2i&1&0&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&-2i&1&\|&0\\ \hidden@noalign{}\hfil\textstyle 0&0&0&-2i&\|&0\end{bmatrix}$		(4.303)
	$\displaystyle\boldsymbol{v}=\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 1% \\ \hidden@noalign{}\hfil\textstyle i\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}=\underbrace{\begin{bmatrix}% \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}}_{\boldsymbol{v_{1}}}+i% \underbrace{\begin{bmatrix}\hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 1\\ \hidden@noalign{}\hfil\textstyle 0\\ \hidden@noalign{}\hfil\textstyle 0\end{bmatrix}}_{\boldsymbol{v}_{2}}$		(4.316)