Answer
See below
Work Step by Step
The general solution for the given differential equation is: $y(x)=c_1e^x+c_2 \sin x+c_3\cos x+A_0e^{-x}$
The trial solution for $y_p= A_0e^{-x}$ can be computed as by plugging back into the given differential equation.
So, we have: $y'''-y''+y'-y=9e^{-x}\\(D^3-D^2+D-1)y_p(x)=9e^{-x}\\ -4A_0e^{-x}=9e^{-x}$
On comparing coefficients, we get: $A_0=-\frac{9}{4}$
Therefore, the general solution for the given differential equation is: $y(x)=c_1e^x+c_2 \sin x+c_3\cos x-\frac{9}{4}e^{-x}$
