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Denote by $I_{n}(x)$ the modified Bessel function of the first kind. Consider the function $g(x)$ given by the Fourier-Neumann type expansion: $$g(x)=\sum_{n=0}^{\infty}a_{n}\frac{I_{n+1}(x)}{x^{n+1}}$$ Where the numbers $a_{n}$ are generated by another series $f(z)$: $$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\;\;\;\;\;\left|z\right|<R$$ Now, we wish for a differential -or integro-differential- equation, satisfied by $g(x)$. I used the fact that: $$\left(\frac{d^{2}}{dx^{2}}+\frac{2n+3}{x}\frac{d}{dx}-1\right)\frac{I_{n+1}(x)}{x^{n+1}}=0$$ After some manipulation, i obtained: $$\left(x\frac{d^{2}}{dx^{2}}+3\frac{d}{dx}-x\right)g(x)=-2\frac{d}{dx}\left(\sum_{n=0}^{\infty}a_{n}n\frac{I_{n+1}(x)}{x^{n+1}}\right)$$ But i wasn't able to relate the RHS to $g(x)$. Hence, the question!

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