solutions to nonhomogeneous system of differential equations with general solution already known Let's say we have the general solution to $X' = A(t)X$, where $X=(x_1, x_2)^T$.
How do you find the general solution to the system $X'= A(t)X + b(t)$
where $b(t)$ is a $2 \times 1$ matrix with two polynomials as entries. How do you find the particular solution?
 A: Let's call the two solutions differently, and clean up notation. Lowercase boldface will denote vectors ($2\times1$). Plain uppercase will denote matrices. Plain lowercase will be scalars. All explicit dependence on $t$ will be dropped. Now the one without any forcing term on the right hand side is denoted with $\mathbf{x}$:
$$
\mathbf{x}' = A\mathbf{x}
$$
The one with forcing we'll denote by $\mathbf{y}$,
$$
\mathbf{y}' = A\mathbf{y} + \mathbf{b}.
$$
Let's guess a form for $\mathbf{y}$. Let's guess it is the product of $\mathbf{x}$ and some unknown scalar function $u$,
$$
\mathbf{y} = u\mathbf{x}
$$
Then your equation becomes
$$
u\mathbf{x}'+u'\mathbf{x} = uA\mathbf{x} + \mathbf{b}
$$
But we already know that $\mathbf{x}'=A\mathbf{x}$ so substitute that
$$
uA\mathbf{x}+u'\mathbf{x} = uA\mathbf{x} + \mathbf{b}
$$
Canceling on both sides we are left with
$$
u'\mathbf{x} =  \mathbf{b}
$$
Multiplying by $\mathbf{x}^T$
$$
u'\mathbf{x}^T\mathbf{x} =  \mathbf{x}^T\mathbf{b}
$$
But $\mathbf{x}^T\mathbf{x}=||\mathbf{x}||^2$, so dividing by that scalar gives
$$
u' = \frac{\mathbf{x}^T\mathbf{b}}{||\mathbf{x}||^2},
$$
and integrating and putting back all the explicit $t$ dependence gives
$$
u(t) = \int\frac{\mathbf{x}^T(t)\mathbf{b}(t)}{||\mathbf{x}(t)||^2}dt
$$
If you take that and multiply this by your original known solution, then I think you have a particular solution $y$.
