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Question Number 207106 by Wuji last updated on 07/May/24

Calculate the generalized solution for the following  system of ODEs:  (dx/dt)=−(1/2)x, (dy/dt)=(1/2)x−(1/4)y,  (dz/dt)=(1/4)y−(1/6)z

$${Calculate}\:{the}\:{generalized}\:{solution}\:{for}\:{the}\:{following} \\ $$$${system}\:{of}\:{ODEs}: \\ $$$$\frac{{dx}}{{dt}}=−\frac{\mathrm{1}}{\mathrm{2}}{x},\:\frac{{dy}}{{dt}}=\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{4}}{y},\:\:\frac{{dz}}{{dt}}=\frac{\mathrm{1}}{\mathrm{4}}{y}−\frac{\mathrm{1}}{\mathrm{6}}{z} \\ $$

Answered by mr W last updated on 07/May/24

 determinant (((−(1/2)−λ),0,0),((1/2),(−(1/4)−λ),0),(0,(1/4),(−(1/6)−λ)))=0  (−(1/2)−λ)(−(1/4)−λ)(−(1/6)−λ)=0  ⇒λ=−(1/2), −(1/4), −(1/6)  V_1 = [((4/3)),((−(8/3))),(1) ]  V_2 = [(0),((−(5/3))),(1) ]  V_3 = [(0),(0),(1) ]   [(x),(y),(z) ]=C_1 e^(−(t/2))  [((4/3)),((−(8/3))),(1) ]+C_2 e^(−(t/4))  [(0),((−(5/3))),(1) ]+C_3 e^(−(t/6))  [(0),(0),(1) ]

$$\begin{vmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}−\lambda}&{\mathrm{0}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{2}}}&{−\frac{\mathrm{1}}{\mathrm{4}}−\lambda}&{\mathrm{0}}\\{\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{4}}}&{−\frac{\mathrm{1}}{\mathrm{6}}−\lambda}\end{vmatrix}=\mathrm{0} \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{2}}−\lambda\right)\left(−\frac{\mathrm{1}}{\mathrm{4}}−\lambda\right)\left(−\frac{\mathrm{1}}{\mathrm{6}}−\lambda\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=−\frac{\mathrm{1}}{\mathrm{2}},\:−\frac{\mathrm{1}}{\mathrm{4}},\:−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${V}_{\mathrm{1}} =\begin{bmatrix}{\frac{\mathrm{4}}{\mathrm{3}}}\\{−\frac{\mathrm{8}}{\mathrm{3}}}\\{\mathrm{1}}\end{bmatrix} \\ $$$${V}_{\mathrm{2}} =\begin{bmatrix}{\mathrm{0}}\\{−\frac{\mathrm{5}}{\mathrm{3}}}\\{\mathrm{1}}\end{bmatrix} \\ $$$${V}_{\mathrm{3}} =\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}={C}_{\mathrm{1}} {e}^{−\frac{{t}}{\mathrm{2}}} \begin{bmatrix}{\frac{\mathrm{4}}{\mathrm{3}}}\\{−\frac{\mathrm{8}}{\mathrm{3}}}\\{\mathrm{1}}\end{bmatrix}+{C}_{\mathrm{2}} {e}^{−\frac{{t}}{\mathrm{4}}} \begin{bmatrix}{\mathrm{0}}\\{−\frac{\mathrm{5}}{\mathrm{3}}}\\{\mathrm{1}}\end{bmatrix}+{C}_{\mathrm{3}} {e}^{−\frac{{t}}{\mathrm{6}}} \begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{1}}\end{bmatrix} \\ $$

Commented by Wuji last updated on 07/May/24

God bless you, sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you},\:\mathrm{sir} \\ $$

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