Answer
$y(x)=c_1e^{x}+c_2 e^{-x}+e^{2x}-e^{3x}$
Work Step by Step
The general solution for the given differential equation is: $y(x)=c_1e^{x}+c_2 e^{-x}+A_0 e^{2x}+B_0e^{3x}$
The trial solution for $y_p= A_0 e^{2x}+B_0e^{3x}$ can be computed as by plugging back into the given differential equation.
So, we have: $(D^2-1)y_p(x)=3e^{2x}-8e^{3x} \\ (D^2-1)(A_0 e^{2x}+B_0e^{3x})=3e^{2x}-8e^{3x} $
On comparing coefficients, we get: $A_0=1$ and $B_0=-1$
Therefore, the general solution for the given differential equation is: $y(x)=c_1e^{x}+c_2 e^{-x}+e^{2x}-e^{3x}$
