Respuesta :

LammettHash
I'm assuming you mean the system is

[tex]\begin{cases}\frac{\mathrm dx}{\mathrm dt}=x+y-z\\\frac{\mathrm dy}{\mathrm dt}=2y\\\frac{\mathrm dz}{\mathrm dt}=y-z\end{cases}[/tex]

If this is correct, then you can immediately solve for [tex]y[/tex], then for [tex]z[/tex], and finally [tex]x[/tex].

From the second ODE, we have

[tex]\dfrac{\mathrm dy}{\mathrm dt}=2y\implies y=C_1e^{2t}[/tex]

Substituting into the third ODE yields

[tex]\dfrac{\mathrm dz}{\mathrm dt}=C_1e^{2t}-z\iff\dfrac{\mathrm dz}{\mathrm dt}+z=C_1e^{2t}[/tex]
[tex]\iff e^t\dfrac{\mathrm d}{\mathrm dt}+e^tz=C_1e^{3t}[/tex]
[tex]\iff\dfrac{\mathrm d}{\mathrm dt}\left[e^tz\right]=C_1e^{3t}[/tex]
[tex]\implies e^tz=\dfrac{C_1}3e^{3t}+C_2[/tex]
[tex]\implies z=\dfrac{C_1}3e^{2t}+C_2e^{-t}[/tex]

Finally, substituting this into the first ODE gives

[tex]\dfrac{\mathrm dx}{\mathrm dt}=x+C_1e^{2t}-\left(\dfrac{C_1}3e^{2t}+C_2e^{-t}\right)\iff\dfrac{\mathrm dx}{\mathrm dt}-x=\dfrac{2C_1}3e^{2t}-C_2e^{-t}[/tex]
[tex]\iff e^{-t}\dfrac{\mathrm dx}{\mathrm dt}-e^{-t}x=\dfrac{2C_1}3e^t-C_2e^{-2t}[/tex]
[tex]\iff\dfrac{\mathrm d}{\mathrm dt}\left[e^{-t}x\right]=\dfrac{2C_1}3e^t-C_2e^{-2t}[/tex]
[tex]\implies e^{-t}x=\dfrac{2C_1}3e^t+\dfrac{C_2}2e^{-2t}+C_3[/tex]
[tex]\implies x=\dfrac{2C_1}3e^{2t}+\dfrac{C_2}2e^{-t}+C_3e^t[/tex]

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