You found the homogeneous solution

x_{h}=c_{1}(-1/sqrt(3),1) e^{-2t} +c_{2} (sqrt(3),1) e^{2t}

To find the inhomogeneous solution you can use one of two methods

A) undetermined coefficients

B) variation of parameters

I will do method A), perhaps somebody else will do method B) for you.

Your inhomogeneous term is (e^{t}, sqrt(3)*e^{-t}), which means x MUST be of the form

x= (A_{1},A_{2}) e^{t} + (B_{1},B_{2}) e^{-t}

We have to find out what A_{1}, A_{2}, B_{1}, and B_{2} are. We do this by substituting x and its derivative

x'=(A_{1},A_{2}) e^{t} - (B_{1},B_{2}) e^{-t}

into the original equation,

x'=(1, sqrt(3), sqrt(3), -1)x+(e^t, sqrt(3)*e^-t)

I will do the matrix term first:

(1, sqrt(3), sqrt(3), -1)x = (1, sqrt(3), sqrt(3), -1) ( (A1,A2) e^{t} + (B1,B2) e^{-t})

= ((A1+sqrt(3)A2)e^{t} + (B1+sqrt(3)B2)e^{-t} , (sqrt(3) A1 -A2)e^{t} + (sqrt(3) B1 -B2)e^{-t} )

Therefore, the first component of x' is

A1 e^{t} - B1 e^{-t} = (A1+sqrt(3)A2)e^{t} + (B1+sqrt(3)B2)e^{-t} +e^{t}

The second component of x' is

A2 e^{t} - B2 e^{-t} = (sqrt(3) A1 -A2)et + (sqrt(3) B1 -B2)e^{-t} +sqrt(3)*e^{-t}

Let's collect all terms with an e^{t} from the first equation and from the second equation:

A1 = A1+sqrt(3)A2 +1

A2 = sqrt(3) A1 -A2

In the first of these, A1 cancels and we get

A2 = -1/sqrt(3).

Plug this into the second equation and get

A1 = -2/3

Now we need to find B1 and B2. For this we collect all terms with an e^{-t} from the first and the second x' equation:

- B1 = B1+sqrt(3)B2

- B2 = sqrt(3) B1 -B2 +sqrt(3)

In the second of these equations, B2 cancels and we get

B1 = -1

Plug this into the first equation and get

B2 = 2/sqrt(3)

Therefore we found

x= (-2/3,-1/sqrt(3)) e^{t} + (-1,2/sqrt(3)) e^{-t}

which is your solution!

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