Question Number 140695 by Chhing last updated on 11/May/21
$$\mathrm{Differential}\:\mathrm{system} \\ $$$$\begin{cases}{\mathrm{x}'+\mathrm{x}−\mathrm{y}−\mathrm{z}=\mathrm{ae}^{\mathrm{2t}} }\\{\mathrm{y}'+\mathrm{y}−\mathrm{z}−\mathrm{x}=\mathrm{be}^{\mathrm{2t}} }\\{\mathrm{z}'+\mathrm{z}−\mathrm{x}−\mathrm{y}=\mathrm{ce}^{\mathrm{2t}} }\end{cases} \\ $$$$\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{constants} \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$
Answered by mr W last updated on 11/May/21
$$\left({x}+{y}+{z}\right)'−\left({x}+{y}+{z}\right)=\left({a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)={Ae}^{{t}} +\left({a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \\ $$$${x}'+\mathrm{2}{x}−\left({x}+{y}+{z}\right)={ae}^{\mathrm{2}{t}} \\ $$$${x}'+\mathrm{2}{x}={Ae}^{{t}} +\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \\ $$$${x}=\frac{\int{e}^{\mathrm{2}{t}} \left({Ae}^{{t}} +\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{2}{t}} \right){dt}+{C}_{\mathrm{1}} }{{e}^{\mathrm{2}{t}} } \\ $$$${x}=\frac{\frac{\mathrm{1}}{\mathrm{4}}{Ae}^{\mathrm{4}{t}} +\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{5}{t}} +{C}_{\mathrm{1}} }{{e}^{\mathrm{2}{t}} } \\ $$$$\Rightarrow{x}=\frac{{A}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +\frac{\left(\mathrm{2}{a}+{b}+{c}\right){e}^{\mathrm{3}{t}} }{\mathrm{5}}+{C}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} \\ $$$$\Rightarrow{y}=\frac{{A}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +\frac{\left({a}+\mathrm{2}{b}+{c}\right){e}^{\mathrm{3}{t}} }{\mathrm{5}}+{C}_{\mathrm{2}} {e}^{−\mathrm{2}{t}} \\ $$$$\Rightarrow{z}=\frac{{A}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +\frac{\left({a}+{b}+\mathrm{2}{c}\right){e}^{\mathrm{3}{t}} }{\mathrm{5}}+{C}_{\mathrm{3}} {e}^{−\mathrm{2}{t}} \\ $$
Commented by Chhing last updated on 12/May/21
$$\mathrm{Thank}\:\mathrm{you} \\ $$