Let $I \subseteq R$ be an interval and $f,g:I \to \mathbb R$ two functions. If there exists a solution $y_\text{part}:I \to \mathbb R$ of the ODE $y'+f(x)y=g(x)$, then the general solution is of the form $y = y_\text{part} + y_\text{hom}$, where $y_\text{hom}$ is the general solution to the homogeneous ODE $y' + f(x)y = 0$.
I know that I need to show that $y = y_\text{part} + y_\text{hom}$ is a solution and then that it is the only solution. However I'm struggling to prove this general theorem, because unlike in excercises I have no concrete formula for $y_\text{part}$, so I can't compute the derivative of $y$ and plug it in the equality to see that it is a solution. I know only that $y_\text{hom}$ is of the form $y_\text{hom} = Ae^{-F(x)}$ where $F(x)$ is an antiderivative of $f(x)$. Can I prove this without specifying the formula for $y_\text{part}$?
We know that the general solution of the homogeneous ODE has the form $$y_\text{hom}(x)=Ay_h(x)=Ae^{-F(x)},\quad A\in\mathbb R.$$ Let $y =Ay_h+y_\text{part}$, then $$ y'+f(x)y= Ay'_h+y'_\text{part}+f(x)\left( Ay_h+y_\text{part} \right) $$ $$ =A\left( y'_h+f(x)y_h \right)+ \left( y'_\text{part}+f(x)y_\text{part} \right) =A\cdot 0+g(x)=g(x), $$ thus, any function of the form $Ay_h+y_\text{part}$ is a solution of the original ODE.
Consider the initial value problem $$ y'(x)+f(x)y=g(x),\quad y(x_0)=y_0. $$ Substituting the initial condition into the solution, one obtains $$ y(x_0)=Ae^{-F(x_0)}+y_\text{part}(x_0)=y_0 $$ and $$\tag{1} A=e^{F(x_0)}\left( y_0-y_\text{part}(x_0)\right). $$ Hence, any solution of the original ODE can be writen in the form $$ y=Ae^{-F(x)}+y_\text{part}(x), $$ where $A$ can be determined from (1).