Let C_{1} = 0, C_{2} = 1, and t = 0, and check. You will get that [x y] = [-1 1]

According to the homogeneous system, then, [x' y'] = [-(-1) + 2 -2(-1) +3(1)] = [3 5]

Yet, as x(t) = -e^{t} + 2te^{t}, x'(t) = -e^{t} + 2te^{t}+ + 2 e^{t} = 2te^{t}+ e^{t} , which means x'(0) = 1. As 1 does not equal 3, this is not a solution.

If, in place of [-1 1], you put a b, you will find, again letting c_{1} = 0 and c_{2} = 1 so that you don't have to deal with so much, that

x(t) = ae^{t }+ 2te^{t}

y(t) = be^{t} + 2te^{t}

x'(t) = ae^{t }+ 2te^{t} + 2e^{t}

y'(t) = be^{t }+ 2te^{t} + 2e^{t}

Let's call these two equations the A equations (for actual)

Yet, from the original equations,

x'(t) = - x(t) + 2y(t)

y'(t) = - 2x(t) + 3y(t)

Substituting,

x'(t) = - (ae^{t }+ 2te^{t}) + 2(be^{t} + 2te^{t}) = 2be^{t} - ae^{t } + 2te^{t}

y'(t) = -2(ae^{t }+ 2te^{t}) + 3(be^{t} + 2te^{t}) = 3be^{t} - 2ae^{t } + 2te^{t}

Let's call these equations the L equations.

The A equations must equal the L equations.

From the 2 x' equations

ae^{t }+ 2te^{t} + 2e^{t} = 2be^{t} - ae^{t } + 2te^{t}

2ae^{t }+ 2e^{t} = 2be^{t}

2a + 2 = b

a + 1 = b

From the 2 y' equations

be^{t }+ 2te^{t} + 2e^{t} = 3be^{t} - 2ae^{t } + 2te^{t}

2ae^{t }+ 2e^{t} = 2be^{t}

2a + 2 = b

a + 1 = b

they in

The x and y solutions are the same. Note that we get b in terms of a. Also, note that they change together as a multiple of [1 1], which not so coincidentally is the left side. Thus,, we have the same general solution.

Arbitrarily, let a = 0 and b = 1

[0 1] replaces the [-1 1] in your solution.

If we check this as we did the original (t = 0 c_{1} = 0, c_{2} = 1) as [x y] = [0 1], from the initial equations, [x' y'] = [-1*0 + 2* 1 -2*0 + 3*1] = [2 3]

As x(t) = 0e^{t} + 2te^{t}, x'(t) = 2te^{t}+ 2e^{t}, so x' (0) = 2

As y(t) = 1e^{t} + 2te^{t}, y'(t) = e^{t }+2te^{t}+ 2e^{t}= 3e^{t }+2te^{t}, so y' (0) = 3

This is [2 3], so our results match.