This is **very **similar to the last post.

Second, you find the particular solution. I will demonstrate how to do this with an example.

# Example

# Example- Repeat Roots

**that**by x as well. Same thing with $e^{-3x}$. That's already in the $y_c$ solution. So, we multiply the $e^{-3x}$ terms by x, but that conflicts with $xe^{-3x}$, so we multiply that by x as well. Our new equation for $y_p$ is $$Ax^2 + Bx + Cxe^{4x} + Dx^2e^{-3x}$$

That's pretty much it. It's very similar to the homogeneous equation, there's a just an additive factor that pushes all of the solutions up from 0 to f(x).