Write
$$\begin{cases}Y(t)- y(t) = C_1 f_1(t) + C_2 f_2(t) + C_3 f_3(t)\\
Y'(t)- y'(t) = C_1 f_1'(t) + C_2 f_2'(t) + C_3 f_3'(t)\\
Y''(t)- y''(t) = C_1 f_1''(t) + C_2 f_2''(t) + C_3 f_3''(t)\\
Y'''(t)- y'''(t) = C_1 f_1'''(t) + C_2 f_2'''(t) + C_3 f_3'''(t).\end{cases}$$
Now you can express that this system has nontrivial solutions for $C$ by the condition
$$\Delta=0$$ which can be developed as
$$(Y(t)- y(t))M_0(t)+(Y'(t)- y'(t))M_1(t)+(Y''(t)- y''(t))M_2(t)+(Y'''(t)- y'''(t))M_2(t)=0$$ (where $M_k$ are minors built on the RHS), a linear ODE.
So you have the following differential equation
$$\frac{\mathrm dx}{\mathrm dt} = c_1 x+ c_2 x^{-1} + c_3$$
We can isolate $x^{-1}$ in the right side of the equation to get
$$\frac{\mathrm dx}{\mathrm dt} = x^{-1}(c_1x^2+c_3x+c_2) =: x^{-1}p(x)$$
We can find the roots of the polynom $p(x):= c_1x^2+c_3x+c_2$ using
$$x_{\pm} = \frac{-c_3\pm\sqrt{c_3^2-4c_1c_2}}{2c_1}$$
So we will have that $p(x) = (x-x_{+})(x - x_{-})$ and we get the differential equation, where $x_{+},x_{-} \in \mathbb{R}$ are constants determined by the constants $c_1,c_2,c_3$ given by the problem with the relation above
$$\frac{\mathrm dx}{\mathrm dt} = x^{-1}(x - x_{+})(x - x_{-}) \implies \frac{x}{(x-x_{+})(x - x_{-})} \mathrm dx = \mathrm dt$$
Now we can use that
$$\frac{x}{(x-x_{+})(x - x_{-})} = \frac{A}{x - x_{+}} + \frac{B}{x - x_{-}}$$
Where we have that
$$A := \frac{-x_{+}}{x_{-} - x_{+}} \,\,\,\,\,\,\,\,B:=\frac{x_{-}}{x_{-}-x_{+}}$$
Now we integrate both sides of the equation, if we put $x(0) := x_0$ we will have
$$A\int_{x_0}^x\frac{\mathrm dx'}{x' - x_{+}} + B\int_{x_0}^x\frac{\mathrm dx'}{x' - x_{-}} = A\ln\left(\frac{x - x_{+}}{x_0 - x_{+}}\right) + B\ln\left(\frac{x - x_{-}}{x_0 - x_{-}}\right) = t$$
Then we'll have that
$$(x - x_{+})^A(x - x_{-})^B = (x_0 - x_{+})^A(x_0 - x_{-})^Be^t$$
You should work out the case $c_3^2 - 4c_1c_2 < 0$.
Best Answer
I guess you mean method of variation of parameters for a third Order Differential Equation. And you already have solved the homogeneous equation.
It's $$\pmatrix { c'_1 \\ c'_2 \\c'_3}=W^{-1}\pmatrix { 0 \\ 0 \\e^{2x} }$$ Where $W^{-1}$ is the inverse of Wronskian matrix and $c_i'$ are derivatives.
Edit I added some calculations
$$W=\pmatrix {e^x & e^{2x} &e^{3x} \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x}}$$ it's easy to find the determinant $|W|=2e^{6x}$ Now you need the inverse of the Wronskian Matrix ...We don't need all the elements of the matrix. Only the last column is needed: $$W^{-1}=\frac 1 {|W|}\pmatrix {* & * &e^{5x} \\ * & *& -2e^{4x} \\ * & * & e^{3x}}$$ So we have that: $$\pmatrix { c'_1 \\ c'_2 \\c'_3}=\frac 1 {|W|}\pmatrix {* & * &e^{5x} \\ * & *& -2e^{4x} \\ * & * & e^{3x}}\pmatrix { 0 \\ 0 \\e^{2x} }$$ $$\pmatrix { c'_1 \\ c'_2 \\c'_3}=\frac 1 {|W|}\pmatrix { e^{7x}\\ -2e^{6x} \\e^{5x} }$$ $$\pmatrix { c'_1 \\ c'_2 \\c'_3}=\frac 1 {2}\pmatrix { e^{x}\\ -2 \\e^{-x} }$$ Now integrate to find the $c_i$ coefficients: $$\pmatrix { c_1 \\ c_2 \\c_3}=\pmatrix { \frac 1 {2}e^{x}\\ -x \\-\frac 1 {2}e^{-x} }$$ The particular solution is therefore: $$y_p=c_1e^x+c_2e^{2x}+c_3e^{3x}$$ $$y_p=\frac 1 {2}e^{2x}-xe^{2x}-\frac 1 {2}e^{2x}$$ $$y_p=-xe^{2x}$$
Variation of Constants is a much better method for your differential equation. Your guess should be: $$y_p=Axe^{2x}$$