Find the general solution of the system x 1 4 3 0 1 6 0 0 2

Find the general solution of the system. x\' = (1 -4 3 0 1 6 0 0 2)x.

Solution

Let, x=[X Y Z]

SO we have

X\'=X

INtegrating gives

X=A e^t

Y=-4X+Y

Y\'-Y=-4Ae^t

(Y\'-Y)e^{-t}=-4A

(Y e^{-t})\'= -4A

Integrating gives

Y e^{4t}=-4At+B

Y=-4 Ate^{-4t}+Be^{-4t}

Z\'=3X+6Y+2Z

Z\'-2Z=3Ae^{t}-24At e^{-4t}+6B e^{-4t}

(Z\'-2Z)e^{-2t}=3Ae^{-t}-24At e^{-6t}+6B e^{-6t}

(Z e^{-2t})\'=3Ae^{-t}-24At e^{-6t}+6B e^{-6t}

Integrating gives

Ze^{-2t}={1/3 e^{-6 t} (A(12t-9e^{5t}+2)-3B)}+C

Z={1/3 e^{-4 t} (A(12te^{2t}-9e^{7t}+2e^{2t})-3Be^{2t})}+Ce^{2t}