So far we have dealt with homogeneous DE, e.g.
$$ y'' - 3y' + 2y = 0 $$When we found the complementary function $y_c(x)$, we got the general solution $y(x)$.
$$ y(x) = y_c(x) = c_1 e^x + c_2 e^{2x} $$Now we have to deal with nonhomogeneous DE, e.g.
$$ y'' - 3y' + 2y = \color{red}{x + \sin x} $$The DE above has right side of $\color{red}{g(x) = x + \sin x}$ or $\color{red}{g(x) \neq 0 \implies}$ DE is not homogeneous.
In this case the general solution consists of complementary function $y_c(x)$ and also of particular solution $y_p(x)$. We showed how to find $y_c(x)$ already in previous lesson. Now remains the problem of finding $\boldsymbol{y_p(x)}$.
There are restrictions to use the methods. The function $g(x)$ has to have members such is $a$, $x^k$, $e^{ax}$, $\sin ax$, $\cos ax$ and their combinations. These functions have finite number of linearly independent derivatives.
For example $x^3$ has a finite number of linearly independent derivatives: $3x^2$, $6x$, $6$. The same can not be said about $x^{-1}$, which has infinite numbers of linearly independent derivatives.
Methods are described further: