Set $$y_p = K \cosh x~, \quad y_p' = K \sinh x~, \quad y_p''=K\cosh x$$ and substitute these functions into the original equation, then: $$K\cosh x+4K\sinh x+4K\cosh x=18\cosh x$$ the coefficient of $\cosh x$ would be $5K=18$, the coefficient of $\sinh x$ is $4K=0$.
(the contradiction would come out ??)
How could I get the particular solution $Y_p$?
Note that $18\cosh(x)=9(e^x+e^{-x})$ contains two exponential terms. This means that you need free parameters for both $$y_p(x)=Ae^x+Be^{-x}.$$ The same functions can also be parametrized using $$ y_p(x)=C\cosh(x)+D\sinh(x). $$ Note that using $A,B$ leads to independent equations, while using $C,D$ leads to a coupled $2\times 2$ linear system, so it is easier to solve the first form and then substitute $e^{\pm x}=\cosh(x)\pm\sinh(x)$.