Question

Alguém me ajudar neste difícil sistema?

Answer 1

Resolver o sistema de equações diferenciais

$\left\{ \!\begin{array}{lc} \dfrac{dy_{1}}{dt}=y_{2}&~~~~\mathbf{(i)}\\\\ \dfrac{dy_{2}}{dt}=y_{3}&~~~~\mathbf{(ii)}\\\\ \dfrac{dy_{3}}{dt}=y_{1}&~~~~\mathbf{(iii)} \end{array} \right.$


sendo $y_1,\;y_2,\;y_3$ funções de $t.$

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Derivando os dois lados em $\mathbf{(i)}$ temos

$\dfrac{d^2 y_{1}}{dt^2}=\dfrac{dy_2}{dt}\\\\\\ \dfrac{d^2 y_{1}}{dt^2}=y_3~~~~\mathbf{(iv)}$


Derivando os dois lados da equação $\mathbf{(iv)}$ acima, temos

$\dfrac{d^3 y_{1}}{dt^3}=\dfrac{dy_3}{dt}\\\\\\ \dfrac{d^3 y_{1}}{dt^3}=y_1\\\\\\ \dfrac{d^3 y_{1}}{dt^3}-y_1=0~~~~~\mathbf{(v)}$


A equação $\mathbf{(v)}$ acima é uma equação diferencial ordinária de 3ª ordem, linear e homogênea.

Polinômio característico:

$\lambda^3-1=0\\\\ \lambda^3=1$


Agora temos que encontrar todas as raízes complexas da equação acima. Ou seja, queremos encontrar $\lambda_1,\;\lambda_2,\;\lambda_3\in\mathbb{C}$ tais que o polinômio característico seja satisfeito.

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Calculando as raízes cúbicas de $1$ pela fórmulas de De Moivre:

Escrevendo $1$ na forma polar (ou trigonométrica)

$1=\cos 0+i\,\mathrm{sen\,}0$


Portanto, as raízes cúbicas de $1$ são

$\lambda_k=1^{1/3}\cdot \left[\cos\!\left(\dfrac{0+k2\pi}{3}\right)+i\,\mathrm{sen}\left(\dfrac{0+k2\pi}{3}\right ) \right ]\\\\\\ \lambda_k=\cos\!\left(\dfrac{k2\pi}{3}\right)+i\,\mathrm{sen}\left(\dfrac{k2\pi}{3}\right )\,,~~~~k=0,\,1,\,2.\\\\\\\\ \bullet~~\lambda_1=\cos 0+i\,\mathrm{sen\,}0\\\\ \boxed{\lambda_1=1}\\\\\\ \bullet~~\lambda_2=\cos\!\left(\dfrac{2\pi}{3}\right)+i\,\mathrm{sen}\left(\dfrac{2\pi}{3}\right )\\\\\\ \boxed{\lambda_2=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\,i}\\\\\\ \bullet~~\lambda_3=\cos\!\left(\dfrac{4\pi}{3}\right)+i\,\mathrm{sen}\left(\dfrac{4\pi}{3}\right )\\\\\\ \boxed{\lambda_3=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\,i}$

(note que $\lambda_2$ e $\lambda_3$ são complexos conjugados)

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A base geradora das soluções da equação homogênea são

$\left\{e^{\lambda_1 t};\;e^{\mathfrak{Re}(\lambda_2)t}\cos\left[\mathfrak{Im}(\lambda_2)\,t \right ];\;e^{\mathfrak{Re}(\lambda_3)\,t}\,\mathrm{sen\,}\left[\mathfrak{Im}(\lambda_3)\,t \right ]\right\}\\\\ \left\{e^{t};\;e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right );\;e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right )\right\}$


Portanto a solução da equação $\mathbf{(v)}$ é

$\boxed{\begin{array}{c} y_1(t)=C_1\,e^{t}+C_2\,e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right )+C_3\,e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right ) \end{array}}$

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Derivando o $y_1(t)$ em relação a $t,$ obtemos

$\dfrac{dy_1}{dt}=C_1\,e^{t}-\dfrac{C_2}{2}\,e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right )-\dfrac{\sqrt{3}\,C_2}{2}\,e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right )\\\\\\ -\dfrac{C_3}{2}\,e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t\right)+\dfrac{\sqrt{3}\,C_3}{2}\,e^{-t/2}\cos \left(\dfrac{\sqrt{3}}{2}\,t \right )\\\\\\\\ \dfrac{dy_1}{dt}=C_1\,e^{t}+\left[-\dfrac{C_2}{2}+\dfrac{\sqrt{3}\,C_3}{2}\right]e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right )-\left[\dfrac{\sqrt{3}\,C_2}{2}+\dfrac{C_3}{2}\right]e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right )\\\\\\ \boxed{\begin{array}{c} y_2(t)=C_1\,e^{t}+\left(-\dfrac{C_2}{2}+\dfrac{\sqrt{3}\,C_3}{2}\right)e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right )-\left(\dfrac{\sqrt{3}\,C_2}{2}+\dfrac{C_3}{2}\right)e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right ) \end{array}}$


De forma análoga, encontra se $y_3(t)$ derivando os dois lados da última igualdade acima:

$\vdots\\\\ \dfrac{dy_2}{dt}=C_1\,e^t-\left(\dfrac{C_2}{2}+\dfrac{\sqrt{3}\,C_3}{2} \right)e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right )+\left(\dfrac{\sqrt{3}\,C_2}{2}+\dfrac{C_3}{2}\right )e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right )\\\\\\ \boxed{\begin{array}{c} y_3(t)=C_1\,e^t-\left(\dfrac{C_2}{2}+\dfrac{\sqrt{3}\,C_3}{2} \right)e^{-t/2}\cos\left(\dfrac{\sqrt{3}}{2}\,t \right )+\left(\dfrac{\sqrt{3}\,C_2}{2}+\dfrac{C_3}{2}\right )e^{-t/2}\,\mathrm{sen}\left(\dfrac{\sqrt{3}}{2}\,t \right ) \end{array}}$