# System of nonhomogenous differential equations - Undetermined coefficients

I am taking a course in differential equations and the book I am using is "Elementary Differential Equations" - E. Boyce & R. DiPrima (tenth edition).

My question is about guessing the form of a particular solution to a non-homogeneous system of equations. Given a non-homogeneous system of differential equations on the form: $x^{\prime} = Px + g(t)$ where $P$ is a constant $n\times n$ - matrix and $g(t)$ is a continuous vector-function for $α \lt t \lt β$.

Let $υ, ν,$ and $\rho$ be $n\times 1$ - vectors.

On page 442 it is said:

The procedure for choosing the form of the solution is substantially the same as that given in Section 3.5 for linear second order equations. The main difference is illustrated by the case of a non-homogeneous term of the form $υe^{\lambda t}$ , where $\lambda$ is a simple root of the characteristic equation. In this situation, rather than assuming a solution of of the form $νte^{\lambda t}$ , it is necessary to use $νte^{\lambda t} + \rho e^{\lambda t}$ , where $ν$ and $ρ$ are determined by substituting into the differential equation.

My question: How did they come up with $νte^{\lambda t} + \rho e^{\lambda t}$ ? I can not see why it is necessary for systems to guess that a particular solution (in this case) is a sum. Why is it not enough to just guess that the particular solution will be on the form $νte^{\lambda t}$?

Hope someone can help me to understand this!

For a scalar differential equation with differential operator $L$, the homogeneous equation is $Lx=0$, so if $Le^{\lambda t}=0$, then $L(\rho e^{\lambda t})=0$ as well.
For the matrix case, the homogeneous equation is $L\vec x=\vec0$, so now if you can find an eigenvalue $\lambda$ and an eigenvector $\vec v$ such that $L\vec ve^{\lambda t}=\vec0$, it's not necessarily true that $L\vec ue^{\lambda t}=\vec0$ for any arbitrary constant vector $\vec u$, only for multiples of that eigenvector $\vec v$. Since $\vec ue^{\lambda t}$ doesn't get wiped out by that differential operator the way $\rho e^{\lambda t}$ did in the scalar case, we will have to consider it in our solution.