Originally Posted by

**Prove It** What you've written is the Characteristic Solution, not the Particular Solution.

Since your RHS of the DE is of the family $\displaystyle \displaystyle e^{4x}$, you would normally choose $\displaystyle \displaystyle Ce^{4x}$ as a particular solution.

But since $\displaystyle \displaystyle e^{4x}$ appears in your Characteristic Solution, you would then normally choose $\displaystyle \displaystyle Cx\,e^{4x}$ as a particular solution.

But since $\displaystyle \displaystyle x\,e^{4x}$ appears in your Characteristic Solution, you would then normally choose $\displaystyle \displaystyle Cx^2e^{4x}$ as a particular solution.

So I agree with the OP's choice of $\displaystyle \displaystyle y_p = Cx^2e^{4x}$.