Does this stand also for $f$?

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We consider the differential equation $Ly=f$ in the ring of exponential sums $\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$ so we have that $f=\sum_{i=0}^n C_i e^{\lambda_i x}$.

If we apply the superposition principle we have to solve differential equations of the form $Ly=e^{bx}$.

If $b$ is a root of the characteristic equation of the homogeneous equation of multiplicity $M$, then the solution is of the form $Cx^Me^{b x}\notin \mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$.

If $b$ is not a root of the characteristic equation of the homogeneous equation, then the solution is of the form $Ce^{b x}\in \mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$.

Therefore, the differential equation has a solution in the ring if $b$ is not a root of the characteristic equation, right?

This is equivalent to $L(e^{b x}) \neq 0$, right?

Does this stand also for the original differential equation? So is it $$L\left (\sum_{i=0}^n C_i e^{\lambda_i x}\right ) \neq 0$$ ?

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If $L=P(D)$, where $P$ is a polynomial, then $L(e^{\lambda_ix})=P(\lambda_i)e^{\lambda_ix}$. Thus, if the $\lambda_i$ avoid the roots of $P$, then for $$ f=\sum_ia_ie^{\lambda_ix} $$ we have $Ly=f$ for $$ y=\sum_i\frac{a_i}{P(\lambda_i)}e^{\lambda_ix} $$