First-order equation
$$y' + p(x)y = q(x) $$The general solution of the corresponding homogeneous equation (written below) is the complementary solution to our original (inhomogeneous) equation:$$ y' + p(x)y = 0$$This homogeneous differential equation can be solved by different methods, for example separation of variables:
$$\frac{d}{dx} y + p(x)y = 0 $$$$\frac{dy}{dx}=-p(x)y $$$${dy \over y} = -{p(x)\,dx},$$$$\int \frac{1}{ y} \, dy = -\int p(x) \, dx $$$$\ln |y| = -\int p(x) \, dx + C $$$$y = \pm e^{-\int p(x) \, dx +C } = C_0 e^{-\int p(x) \, dx}$$The complementary solution to our original equation is therefore:$$y_c = C_0 e^{-\int p(x) \, dx}$$Now we return to solving the non-homogeneous equation$$y' + p(x)y = q(x)$$Using the method variation of parameters, the particular solution is formed by multiplying the complementary solution by an unknown function $C(x)$$$y_p = C(x) e^{-\int p(x) \, dx}$$By substituting the particular solution into the non-homogeneous equation, we can find $C(x)$$$ C' (x) e^{-\int p(x) \, dx} - C(x) p(x) e^{-\int p(x) \, dx} + p(x) C(x) e^{-\int p(x) \, dx} = q(x)$$$$ C' (x) e^{-\int p(x) \, dx} = q(x)$$$$ C' (x) = q(x) e^{\int p(x) \, dx}$$$$C(x) =\int q(x) e^{\int p(x) \, dx} \, dx + C_1$$We only need a single particular solution, so we arbitrarily select $C_1=0$ for simplicity. Therefore the particular solution is$$y_p =e^{-\int p(x) \, dx} \int q(x) e^{\int p(x) \, dx} \, dx$$The final solution of the differential equation is$$\begin{align*}y &= y_c + y_p\\&=C_0 e^{-\int p(x) \, dx} + e^{-\int p(x) \, dx} \int q(x) e^{\int p(x) \, dx} \, dx\end{align*}$$This recreates the method of integrating factors.

Specific second-order equation
Let us solve$$ y''+4y'+4y = \cosh x$$We want to find the general solution to the differential equation, that is, we want to find solutions to the homogeneous differential equation$$y''+4y'+4y=0.$$The characteristic equation is:$$\lambda^2+4\lambda+4=(\lambda+2)^2=0 $$Since $\lambda=-2$ is a repeated root, we have to introduce a factor of $x$ for one solution to ensure linear independence: $u_1=e^{−2x}$ and $u_2=xe^{−2x}$. The Wronskian of these two functions is$$W=\begin{vmatrix}e^{-2x} & xe^{-2x} \\-2e^{-2x} & -e^{-2x}(2x-1)\\\end{vmatrix} = -e^{-2x}e^{-2x}(2x-1)+2xe^{-2x}e^{-2x} = e^{-4x}. $$Because the Wronskian is non-zero, the two functions are linearly independent, so this is in fact the general solution for the homogeneous differential equation (and not a mere subset of it).
We seek functions $A(x)$ and $B(x)$ so $A(x)u_1+B(x)u_2$ is a particular solution of the non-homogeneous equation. We need only calculate the integrals$$A(x) = - \int {1\over W} u_2(x) b(x)\,\mathrm dx,\; B(x) = \int {1 \over W} u_1(x)b(x)\,\mathrm dx$$Recall that for this example$$b(x) = \cosh x$$That is,$$A(x) = - \int {1\over e^{-4x}} xe^{-2x} \cosh x \,\mathrm dx = - \int xe^{2x}\cosh x \,\mathrm dx = -{1\over 18}e^x\left(9(x-1)+e^{2x}(3x-1)\right)+C_1$$$$B(x) = \int {1 \over e^{-4x}} e^{-2x} \cosh x \,\mathrm dx = \int e^{2x}\cosh x\,\mathrm dx ={1\over 6}e^x\left(3+e^{2x}\right)+C_2 $$where $C_1$ and $C_2$ are constants of integration.

General second-order equation
We have a differential equation of the form$$u''+p(x)u'+q(x)u=f(x)$$and we define the linear operator$$L=D^2+p(x)D+q(x)$$where $D$ represents the differential operator.  We therefore have to solve the equation $L u(x)=f(x)$ for $u(x)$, where $L$ and $f(x)$ are known.
We must solve first the corresponding homogeneous equation:$$u''+p(x)u'+q(x)u=0$$by the technique of our choice. Once we've obtained two linearly independent solutions to this homogeneous differential equation (because this ODE is second-order) — call them $u_1$ and $u_2$ — we can proceed with variation of parameters.
Now, we seek the general solution to the differential equation $u_G(x)$ which we assume to be of the form$$u_G(x)=A(x)u_1(x)+B(x)u_2(x).$$Here, $A(x)$ and $B(x)$ are unknown and $u_1(x)$ and $u_2(x)$ are the solutions to the homogeneous equation.  (Observe that if $A(x)$ and $B(x)$ are constants, then $Lu_G(x)=0$.)  Since the above is only one equation and we have two unknown functions, it is reasonable to impose a second condition. We choose the following:$$A'(x)u_1(x)+B'(x)u_2(x)=0.$$Now,$$\begin{align*}
u_G'(x) &= \left (A(x)u_1(x)+B(x)u_2(x) \right )' \\
&= \left (A(x)u_1(x) \right )'+ \left (B(x)u_2(x) \right )'\\
&=A'(x)u_1(x)+B'(x)u_2(x)+A(x)u_1'(x)+B(x)u_2'(x) \\
&= A(x)u_1'(x)+B(x)u_2'(x)
\end{align*}$$Differentiating again (omitting intermediary steps)$$u_G''(x)=A(x)u_1''(x)+B(x)u_2''(x)+A'(x)u_1'(x)+B'(x)u_2'(x).$$Now we can write the action of $L$ upon $u_G$ as$$Lu_G=A(x)Lu_1(x)+B(x)Lu_2(x)+A'(x)u_1'(x)+B'(x)u_2'(x).$$Since $u_1$ and $u_2$ are solutions, then$$Lu_G=A'(x)u_1'(x)+B'(x)u_2'(x).$$We have the system of equations$$\begin{bmatrix}
u_1(x)  & u_2(x) \\
u_1'(x) & u_2'(x) \end{bmatrix}
A'(x) \\
B'(x)\end{bmatrix} =
\begin{bmatrix} 0 \\ f \end{bmatrix}.$$Expanding,$$\begin{bmatrix}
A'(x)u_1'(x)+B'(x)u_2'(x) \end{bmatrix}
= \begin{bmatrix} 0\\f\end{bmatrix}.$$So the above system determines precisely the conditions$$A'(x)u_1(x)+B'(x)u_2(x)=0.$$$$A'(x)u_1'(x)+B'(x)u_2'(x)=Lu_G=f.$$We seek $A(x)$ and $B(x)$ from these conditions, so, given$$\begin{bmatrix}u_1(x)  & u_2(x) \\
u_1'(x) & u_2'(x)
A'(x) \\
B'(x)\end{bmatrix} =
f\end{bmatrix}$$we can solve for $(A'(x), B'(x))^{\top}$, so$$\begin{bmatrix} A'(x) \\ B'(x) \end{bmatrix} =
u_1(x)  & u_2(x) \\
u_1'(x) & u_2'(x)
\begin{bmatrix} 0\\ f \end{bmatrix} =\frac{1}{W} \begin{bmatrix}
u_2'(x)  & -u_2(x) \\
-u_1'(x) & u_1(x) \end{bmatrix}
\begin{bmatrix} 0\\ f \end{bmatrix},$$where $W$ denotes the Wronskian of $u_1$ and $u_2$. (We know that $W$ is nonzero, from the assumption that $u_1$ and $u_2$ are linearly independent.) So,$$ \begin{align*}A'(x) &= - {1\over W} u_2(x) f(x), & B'(x) &= {1 \over W} u_1(x)f(x) \\A(x)  &= - \int {1\over W} u_2(x) f(x)\,\mathrm dx, & B(x) &= \int {1 \over W} u_1(x)f(x)\,\mathrm dx\end{align*}$$While homogeneous equations are relatively easy to solve, this method allows the calculation of the coefficients of the general solution of the inhomogeneous equation, and thus the complete general solution of the inhomogeneous equation can be determined.
Note that $A(x)$ and $ B(x)$ are each determined only up to an arbitrary additive constant (the constant of integration). Adding a constant to $A(x)$ or $B(x)$ does not change the value of $Lu_G(x)$ because the extra term is just a linear combination of $u_1$ and $u_2$, which is a solution of $L$ by definition.

Description of method
Given an ordinary non-homogeneous linear differential equation of order $n$$$y^{(n)}(x) + \sum_{i=0}^{n-1} a_i(x) y^{(i)}(x) = b(x).\label1\tag1$$Let $y_1(x), \ldots, y_n(x)$ be a fundamental system of solutions of the corresponding homogeneous equation$$y^{(n)}(x) + \sum_{i=0}^{n-1} a_i(x) y^{(i)}(x) = 0.\label2\tag2$$Then a particular solution to the non-homogeneous equation is given by$$y_p(x) = \sum_{i=1}^{n} c_i(x) y_i(x)\label3\tag3$$where the $c_i(x)$ are differentiable functions which are assumed to satisfy the conditions$$\sum_{i=1}^n c_i'(x) y_i^{(j)}(x) = 0, \quad j = 0,\ldots, n-2.\label4\tag4$$Starting with \eqref{3}, repeated differentiation combined with repeated use of \eqref{4} gives$$y_p^{(j)}(x) = \sum_{i=1}^{n} c_i(x) y_i^{(j)}(x), \quad j=0,\ldots,n-1 \, .\label5\tag5$$One last differentiation gives$$y_p^{(n)}(x)=\sum_{i=1}^n c_i'(x)y_i^{(n-1)}(x)+\sum_{i=1}^n c_i(x) y_i^{(n)}(x) \, .\label6\tag6$$By substituting \eqref{3} into \eqref{1} and applying \eqref{5} and \eqref{6} it follows that$$\sum_{i=1}^n c_i'(x) y_i^{(n-1)}(x) = b(x).\tag7\label7$$The linear system \eqref{4} and \eqref{7} of $n$ equations can then be solved using Cramer's rule yielding$$c_i'(x) = \frac{W_i(x)}{W(x)}, \, \quad i=1,\ldots,n$$where $W(x)$ is the Wronskian determinant of the fundamental system and $W_i(x)$ is the Wronskian determinant of the fundamental system with the $i$-th column replaced by $(0, 0, \ldots, b(x))$.The particular solution to the non-homogeneous equation can then be written as$$\sum_{i=1}^n y_i(x) \, \int \frac{W_i(x)}{W(x)}\, \mathrm dx.$$