Consider the equation \begin{equation} \frac{f(p)}{f(s_{1})}\mathcal{L}(y)(s_{1})+\frac{g(p)}{g(s_{2})}\mathcal{L}% (y)(s_{2})=\mathcal{L(}y)\mathbf{(}p), \end{equation} where $\mathcal{L}$ is the Laplace transform of $y$ which we assume well defined and $f,g$ are two functions such that \begin{equation*} f(p)=0\text{ }\Longleftrightarrow p=s_{2},\text{ }g(p)=0\text{ }% \Longleftrightarrow p=s_{1}. \end{equation*} I want to prove that such a function $y$ exists if and only if $p=s_{1}$ or $% p=s_{2}.$ The sufficient part is clear; for $p=s_{1}$ we get \begin{eqnarray*} \frac{f(s_{1})}{f(s_{1})}\mathcal{L}(y)(s_{1})+0 &=&\mathcal{L(}y)\mathbf{(}% s_{1}), \\ &\Longrightarrow &\mathcal{L}(y)(s_{1})=\mathcal{L}(y)(s_{1}). \end{eqnarray*}

The same thing holds for $p=s_{2}.$ I want to prove that a solution $y$ exists if and only if $p=s_{1}$ or $p=s_{2}.$ The sufficient part is clear, but I cannot prove the necessary part. Any ideas?. Thank you.