6.36 Example: exceptional case

In this case the auxiliary equation is $s^{2}+s-2={(s+2)(s-1)}$ so the CF is $Ae^{x}+Be^{-2x}$. Now the function $q(x)$ on the right-hand side is already a solution for the homogeneous equation, so to find the PI we consider $y$ of the form $Cxe^{-2x}$. Then we have $y^{\prime}={Ce^{-2x}-2Cxe^{-2x}}$ and $y^{\prime\prime}={4Cxe^{-2x}-4Ce^{-2x}.}$ Grouping terms together, we obtain

and hence the PI is: $\frac{-1}{3}xe^{-2x}$.