Question

Resolva a equação ordinária

X" + 2x= cos(√2)t

Answer 1

$x''+2x=\cos\!\big(\sqrt{2}\,t\big)$


Equação diferencial ordinária de 2ª ordem, linear, não-homogênea e a coeficientes constantes.


Resolvendo a equação característica:

$\lambda^2+2=0\\\\ \lambda^2=-2\\\\ \lambda^2=2\cdot (-1)\\\\ \lambda=\pm \sqrt{2}\,i~~~~\big(\text{ou ainda }\lambda=0\pm \sqrt{2}\,i\big)$


As raízes da equação característica são complexos conjugados:

$\lambda=\alpha\pm \beta i$

onde a $\alpha=0$ e $\beta=\sqrt{2}.$


Base geradora da solução da EDO homogênea:

$\big\{e^{\alpha t}\cos \beta t,\,e^{\alpha t}\mathrm{sen\,} \beta t\big\}\\\\\\ \left\{e^{0 t}\cos\!\big(\sqrt{2}\,t\big),\,e^{0 t}\mathrm{sen}\big(\sqrt{2}\,t\big)\right\}\\\\\\ \left\{\cos\!\big(\sqrt{2}\,t\big),\,\mathrm{sen}\big(\sqrt{2}\,t\big)\right\}$


Solução da EDO homogênea:

$x_h(t)=C_1\cos\!\big(\sqrt{2}\,t\big)+C_2\,\mathrm{sen}\big(\sqrt{2}\,t\big)$

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Solução particular para a EDO não-homogênea:

(método dos coeficientes a determinar)

$x_p(t)=At\cos\!\big(\sqrt{2}\,t\big)+Bt\,\mathrm{sen}\big(\sqrt{2}\,t\big)$


Derivando em relação a $t,$

$x_p'=\left[At\cos\!\big(\sqrt{2}\,t\big)+Bt\,\mathrm{sen}\big(\sqrt{2}\,t\big) \right ]'\\\\\\ x_p'=A\cos\!\big(\sqrt{2}\,t\big)-\sqrt{2}\,At\,\mathrm{sen}\big(\sqrt{2}\,t\big)+B\,\mathrm{sen}\big(\sqrt{2}\,t\big)+\sqrt{2}\,Bt\cos\!\big(\sqrt{2}\,t\big)\\\\\\ x_p'=\big(A+\sqrt{2}\,Bt\big)\cos\!\big(\sqrt{2}\,t\big)+\big(\!\!-\sqrt{2}\,At+B\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)$


Derivando novamente em relação a $t,$

$x_p''=\left[\big(A+\sqrt{2}\,Bt\big)\cos\!\big(\sqrt{2}\,t\big)+\big(\!\!-\sqrt{2}\,At+B\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)\right ]'\\\\\\\\ \begin{array}{ll} x_p''&=\sqrt{2}\,B\cos\big(\sqrt{2}\,t\big)+\big(A+\sqrt{2}\,Bt\big)\cdot \big(\!\!-\!\sqrt{2}\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)\\\\ &-\sqrt{2}\,A\,\mathrm{sen}\big(\sqrt{2}\,t\big)+\big(\!\!-\!\sqrt{2}\,At+B\big)\cdot \sqrt{2}\cos \big(\sqrt{2}\,t\big) \end{array}\\\\\\\\ \begin{array}{ll} x_p''&=\sqrt{2}\,B\cos\big(\sqrt{2}\,t\big)+\big(\!\!-\!\sqrt{2}\,A-2Bt\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)\\\\ &-\sqrt{2}\,A\,\mathrm{sen}\big(\sqrt{2}\,t\big)+\big(\!\!-2At+\sqrt{2}\,B\big)\cos \big(\sqrt{2}\,t\big) \end{array}$


$x_p''=\big(\!\!-2At+2\sqrt{2}\,B\big)\cos\big(\sqrt{2}\,t\big)+\big(\!\!-2\sqrt{2}A-2Bt\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)$


Substituindo na EDO não-homogênea, devemos ter

$x_p''+2x_p=\cos\big(\sqrt{2}\,t\big)\\\\\\ \begin{array}{ll} \big(\!\!-2At+2\sqrt{2}\,B\big)\cos\big(\sqrt{2}\,t\big)+\big(\!\!-2\sqrt{2}A-2Bt\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)\\\\ +2\left[At\cos\!\big(\sqrt{2}\,t\big)+Bt\,\mathrm{sen}\big(\sqrt{2}\,t\big) \right ]=\cos\big(\sqrt{2}\,t\big) \end{array}\\\\\\\\ \begin{array}{ll} \big(\!\!-2At+2\sqrt{2}\,B\big)\cos\big(\sqrt{2}\,t\big)+\big(\!\!-2\sqrt{2}A-2Bt\big)\,\mathrm{sen}\big(\sqrt{2}\,t\big)\\\\ +2At\cos\!\big(\sqrt{2}\,t\big)+2Bt\,\mathrm{sen}\big(\sqrt{2}\,t\big) =\cos\big(\sqrt{2}\,t\big) \end{array}$


Simplificando o lado esquerdo da igualdade acima, ficamos com

$2\sqrt{2}\,B\cos\big(\sqrt{2}\,t\big)-2\sqrt{2}\,A\,\mathrm{sen}\big(\sqrt{2}\,t\big)=\cos\big(\sqrt{2}\,t\big)$


Daqui, obtemos o seguinte sistema:

$\left\{ \!\begin{array}{lcl} 2\sqrt{2}\,B=1&~\Rightarrow~&B=\dfrac{1}{2\sqrt{2}}\\\\ -2\sqrt{2}\,A=0&~\Rightarrow~&A=0 \end{array} \right.$


Então, a solução particular para a EDO não-homogênea é

$x_p(t)=\dfrac{1}{2\sqrt{2}}\,t\,\mathrm{sen}\big(\sqrt{2}\,t\big)$

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Solução geral para a EDO dada inicialmente:

$x(t)=x_h(t)+x_p(t)\\\\ \boxed{\begin{array}{c}x(t)=C_1\cos\!\big(\sqrt{2}\,t\big)+C_2\,\mathrm{sen}\big(\sqrt{2}\,t\big)+\dfrac{1}{2\sqrt{2}}\,t\,\mathrm{sen}\big(\sqrt{2}\,t\big) \end{array}}$


com $C_1,\,C_2\in\mathbb{R}.$


Bons estudos! :-)