Question Number 180718 by a.lgnaoui last updated on 16/Nov/22
$${Resoudre}\:{le}\:{systeme} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{4}{x}+\mathrm{6}{y} \\ $$$$\frac{{dy}}{{dt}}=−\mathrm{3}{x}−\mathrm{5}{y} \\ $$$$\frac{{dz}}{{dt}}=−\mathrm{3}{x}−\mathrm{6}{y}−\mathrm{5}{z} \\ $$
Answered by mr W last updated on 16/Nov/22
![determinant (((4−λ),6),((−3),(−5−λ)))=0 −(4−λ)(5+λ)+3×6=0 (λ+2)(λ−1)=0 ⇒λ=−2, 1 V_1 = [(1),((−1)) ] V_2 = [(2),((−1)) ] ⇒x=c_1 e^(−2t) +2c_2 e^t ⇒y=−c_1 e^(−2t) −c_2 e^t (dz/dt)+5z=3c_1 e^(−2t) ⇒z=((∫3c_1 e^(−2t) e^(5t) dt)/e^(5t) )+c_3 ⇒z=c_1 e^(−2t) +c_3](https://www.tinkutara.com/question/Q180761.png)
$$\begin{vmatrix}{\mathrm{4}−\lambda}&{\mathrm{6}}\\{−\mathrm{3}}&{−\mathrm{5}−\lambda}\end{vmatrix}=\mathrm{0} \\ $$$$−\left(\mathrm{4}−\lambda\right)\left(\mathrm{5}+\lambda\right)+\mathrm{3}×\mathrm{6}=\mathrm{0} \\ $$$$\left(\lambda+\mathrm{2}\right)\left(\lambda−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=−\mathrm{2},\:\mathrm{1} \\ $$$${V}_{\mathrm{1}} =\begin{bmatrix}{\mathrm{1}}\\{−\mathrm{1}}\end{bmatrix} \\ $$$${V}_{\mathrm{2}} =\begin{bmatrix}{\mathrm{2}}\\{−\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow{x}={c}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} +\mathrm{2}{c}_{\mathrm{2}} {e}^{{t}} \\ $$$$\Rightarrow{y}=−{c}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} −{c}_{\mathrm{2}} {e}^{{t}} \\ $$$$\frac{{dz}}{{dt}}+\mathrm{5}{z}=\mathrm{3}{c}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} \\ $$$$\Rightarrow{z}=\frac{\int\mathrm{3}{c}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} {e}^{\mathrm{5}{t}} {dt}}{{e}^{\mathrm{5}{t}} }+{c}_{\mathrm{3}} \\ $$$$\Rightarrow{z}={c}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} +{c}_{\mathrm{3}} \\ $$
Commented by a.lgnaoui last updated on 24/Nov/22
$${thank}\:{you}\:{very}\:{much} \\ $$