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

$$\mathrm{Differential}\mathrm{system}$$$$\{\begin{array}{l}{\mathrm{x}}^{\prime }+\mathrm{x}-\mathrm{y}-\mathrm{z}={\mathrm{ae}}^{2\mathrm{t}}\\ {\mathrm{y}}^{\prime }+\mathrm{y}-\mathrm{z}-\mathrm{x}={\mathrm{be}}^{2\mathrm{t}}\\ {\mathrm{z}}^{\prime }+\mathrm{z}-\mathrm{x}-\mathrm{y}={\mathrm{ce}}^{2\mathrm{t}}\end{array}$$$$\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}\mathrm{constants}$$$$\mathrm{help}\mathrm{me}$$

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

$${(x+y+z)}^{\prime }-(x+y+z)=(a+b+c)e^{2t}$$$$\Rightarrow (x+y+z)={Ae}^t+(a+b+c)e^{2t}$$$$x^{\prime }+2x-(x+y+z)={ae}^{2t}$$$$x^{\prime }+2x={Ae}^t+(2a+b+c)e^{2t}$$$$x=\frac{\int e^{2t}({Ae}^t+(2a+b+c)e^{2t})dt+C_1}{e^{2t}}$$$$x=\frac{\frac{1}{4}{Ae}^{4t}+\frac{1}{5}(2a+b+c)e^{5t}+C_1}{e^{2t}}$$$$\Rightarrow x=\frac{A}{4}{e}^{2t}+\frac{(2a+b+c)e^{3t}}{5}+C_1{e}^{-2t}$$$$\Rightarrow y=\frac{A}{4}{e}^{2t}+\frac{(a+2b+c)e^{3t}}{5}+C_2{e}^{-2t}$$$$\Rightarrow z=\frac{A}{4}{e}^{2t}+\frac{(a+b+2c)e^{3t}}{5}+C_3{e}^{-2t}$$

Commented by Chhing last updated on 12/May/21

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

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