Let's say that we have $n$ differential equations written in the form: $x'(t) = Ax(t) + v \exp(\lambda t)$, where $v$ is the eigenvector of $A$ such that $A v = \lambda v$ and $A$ is a $n \times n$ matrix. Assume that $A$ has $n$ linear independent eigenvectors (and $n$ pairwise different eigenvalues).
Show that there is no vector $a \in \mathbb R^n$ such that $\psi(t) = a \exp(\lambda t)$ is a solution, but $\psi(t) = a\exp(\lambda t) + bt\exp(\lambda t)$ for some values $a, b \in \mathbb R^n$ is a solution.
I'm not sure if I have to show that every solution of my system is of the given form OR if there is at least one solution (for particular $a$ and $b$).
What have I tried?
$$a\lambda \exp(\lambda t) + b \exp(\lambda t) + b t \lambda \exp(\lambda t) = A(a\exp(\lambda t) + bt\exp(\lambda t))+ v \exp(\lambda t).$$ This is equivalent to $$(a \lambda + b) + t \cdot b\lambda = (Aa + v) + t \cdot Ab,$$ so $Ab = b\lambda$ (and this implies $b = k \cdot v$ for real $k$) and $a \lambda + b = Aa + v$, but how to proceed now?
You have: $(A-\lambda)a = (k-1)v$.
Now, $\psi(t) = a \exp(\lambda t)$, $\psi'(t) = a \lambda \exp(\lambda t)$, we have: $ \lambda a \ne Aa + v \ \forall a \in \mathbb{R}^n$, so for any vector $a\in R^n$:
$(A-\lambda)a \ne -v$, so $k-1\ne -1$, i.e $k\ne 0$.
So you found yourself suitable $a,b$, $b =kv , k \ne 0$, $a$ satisfies: $(A-\lambda)a = (k-1)v$.