How should I solve this differential equation with f(x) not given? 
I mean I cannot evaluate the particular integral since f(x) is not known. And how do I come up with double integral formula shown above? 
 A: Note that 
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2 \frac{\mathrm{d}y}{\mathrm{d}x} + y = \left( \frac{\mathrm{d}}{\mathrm{d}x} - \operatorname{id} \right)^2 y.$$
So first we solve the equation
$$\frac{\mathrm{d}y}{\mathrm{d}x} - y = f(x).$$
For that we multiply both sides by $e^{-x}$ and contract
$$
\begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x} \cdot e^{-x} - y e^{-x} & = e^{-x} f(x) \\ 
\left( y e^{-x} \right)' & = e^{-x} f(x) \\
y e^{-x} & = \int e^{-x} f(x) \, \mathrm{d} x \\
y & = e^x \int e^{-x} f(x) \, \mathrm{d} x
\end{align*}
$$
We denote the solution by $y_1$. Now by the observation in the beginning, we have
$$\left( \frac{\mathrm{d}}{\mathrm{d}x} - \operatorname{id} \right)\left( \frac{\mathrm{d}}{\mathrm{d}x} - \operatorname{id} \right) y = f(x)$$
hence
$$\left( \frac{\mathrm{d}}{\mathrm{d}x} - \operatorname{id} \right) y = y_1$$
which we solve in the same way, obtaining
$$y = e^x \int e^{-x} y_1 \, \mathrm{d} x.$$
Substituting $y_1$ yields
$$y = e^x \int e^{-x} \left[ e^x \int e^{-x} f(x) \, \mathrm{d} x \right] \, \mathrm{d} x = e^x \iint e^{-x} f(x) \, \mathrm{d} x.$$
A: This can be solved using variation of parameters
The homogeneous equation
$$ y'' - 2y' + y = 0 $$
has solutions $y_1 = e^x$, $y_2 = xe^x$
Let the particular solution have the form
$$ y_p(x) = v_1(x)e^x + v_2(x)xe^x $$
Applying the general method for second order equations we can impose the conditions
$$ {v_1}'e^x + {v_2}'xe^x = 0 $$
$$ {v_1}'e^x + {v_2}'(e^x + xe^x) = f(x) $$
Solving this gives
$$ {v_1}' = -xe^{-x}f(x) $$
$$ {v_2}' = e^{-x}f(x) $$
So the particular solution is 
$$ y_p = -e^x\int xe^{-x}f(x)dx + xe^x\int e^{-x}f(x) dx $$
You can use integration by parts to prove
$$ \int xe^{-x}f(x) dx = x \int e^{-x}f(x) - \int\int e^{-x}f(x) dxdx $$
Which gives the final form
$$ y_p = e^x \int\int e^{-x} f(x)dxdx$$
