`(dx)/(dt) = x - 2y`

`(dy)/(dt) = 2x+y`

Now for a system of ode,

`(dx)/(dt) = ax+by`

`(dy)/(dt) = cx+dy`

The characteristic equation is given by,

`lambda^2-(a+d)lambda+ad-bc = 0`

If the solutions of the characterstic equation are lambda_1 and lambda_2 respectively.

Then, the general solution of the system of ode is given by,

`x = C_1be^(lambda_1t) + C_2be^(lambda_2t)`

`y = C_1(lambda_1 - a)e^(lambda_1t) + C_2(lambda_2 - a)e^(lambda_2t)`

Where C1 and C2 are arbitrary constants.

In our example, a = 1,b=-2, c= 2 and d=1.

Therefore the characteristic equation is,

`lambda^2-(1+1)lambda+1+4 = 0`

`lambda^2-2lambda+1 = -4`

`(lambda-1)^2 = (2i)^2`

`lambda-1 = +-2i`

`lambda = 1+-2i`

When the roots are complex `(lambda=sigma+-betai)` , the general solution is given by,

`x = be^(sigmat)[C1 sin(betat) + C2 cos(betat)]`

`y = e^(sigmat){[(sigma - a)C1 - betaC2] sin(betat) + [betaC1 + (sigma - a)C2]cos(betat)}`

Where C1 and C2 are arbitrary constants.

For our example, `a=1` , `b= -2` , `sigma =1` and `beta =2` .

`x = -2e^(t)(C1 sin(2t) + C2 cos(2t))`

`y = e^t{[(1 - 1)C1 - 2C2] sin(2t) + [2C1 + (1-1)C2]cos(2t)}`

`y = e^t{-2C2sin(2t) + 2C1cos(2t)}`

`y = 2e^t(-C2sin(2t) + C1cos(2t))`

The general solution is,

`x = -2e^(t)(C1 sin(2t) + C2 cos(2t))`

`y = 2e^t(-C2sin(2t) + C1cos(2t))`

Where C1 and C2 are arbitrary constants.