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Question Number 140695 by Chhing last updated on 11/May/21

Differential system { ((x′+x−y−z=ae^(2t) )),((y′+y−z−x=be^(2t) )),((z′+z−x−y=ce^(2t) )) :} a,b,c are constants help me

$$\mathrm{Differential}\:\mathrm{system} \\ $$$$\begin{cases}{\mathrm{x}'+\mathrm{x}−\mathrm{y}−\mathrm{z}=\mathrm{ae}^{\mathrm{2t}} }\\{\mathrm{y}'+\mathrm{y}−\mathrm{z}−\mathrm{x}=\mathrm{be}^{\mathrm{2t}} }\\{\mathrm{z}'+\mathrm{z}−\mathrm{x}−\mathrm{y}=\mathrm{ce}^{\mathrm{2t}} }\end{cases} \\ $$$$\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{constants} \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$

Answered by mr W last updated on 11/May/21

(x+y+z)′−(x+y+z)=(a+b+c)e^(2t) ⇒(x+y+z)=Ae^t +(a+b+c)e^(2t) x′+2x−(x+y+z)=ae^(2t) x′+2x=Ae^t +(2a+b+c)e^(2t) x=((∫e^(2t) (Ae^t +(2a+b+c)e^(2t) )dt+C_1 )/e^(2t) ) x=(((1/4)Ae^(4t) +(1/5)(2a+b+c)e^(5t) +C_1 )/e^(2t) ) ⇒x=(A/4)e^(2t) +(((2a+b+c)e^(3t) )/5)+C_1 e^(−2t) ⇒y=(A/4)e^(2t) +(((a+2b+c)e^(3t) )/5)+C_2 e^(−2t) ⇒z=(A/4)e^(2t) +(((a+b+2c)e^(3t) )/5)+C_3 e^(−2t)

$$\left({x}+{y}+{z}\right)'−\left({x}+{y}+{z}\right)=\left({a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)={Ae}^{{t}} +\left({a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \\ $$$${x}'+\mathrm{2}{x}−\left({x}+{y}+{z}\right)={ae}^{\mathrm{2}{t}} \\ $$$${x}'+\mathrm{2}{x}={Ae}^{{t}} +\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \\ $$$${x}=\frac{\int{e}^{\mathrm{2}{t}} \left({Ae}^{{t}} +\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \right){dt}+{C}_{\mathrm{1}} }{{e}^{\mathrm{2}{t}} } \\ $$$${x}=\frac{\frac{\mathrm{1}}{\mathrm{4}}{Ae}^{\mathrm{4}{t}} +\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{5}{t}} +{C}_{\mathrm{1}} }{{e}^{\mathrm{2}{t}} } \\ $$$$\Rightarrow{x}=\frac{{A}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +\frac{\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{3}{t}} }{\mathrm{5}}+{C}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} \\ $$$$\Rightarrow{y}=\frac{{A}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +\frac{\left({a}+\mathrm{2}{b}+{c}\right){e}^{\mathrm{3}{t}} }{\mathrm{5}}+{C}_{\mathrm{2}} {e}^{−\mathrm{2}{t}} \\ $$$$\Rightarrow{z}=\frac{{A}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +\frac{\left({a}+{b}+\mathrm{2}{c}\right){e}^{\mathrm{3}{t}} }{\mathrm{5}}+{C}_{\mathrm{3}} {e}^{−\mathrm{2}{t}} \\ $$

Commented by Chhing last updated on 12/May/21

Thank you

$$\mathrm{Thank}\:\mathrm{you} \\ $$

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