Is there a general exact solution for a Normal distribution convolved to $e^{y(x)}$, where $y(x)$ is a polynomial?
$$e^{y(x)}*N(x,v)=\int_{-\infty}^{\infty}e^{y(z)}N(x-z,v)dz$$ $$y(x)\equiv\sum_{n=0}^{m}c_{n}x^{n}$$ $$N(x,v)\equiv\frac{1}{\sqrt{2\pi v}}e^{-\frac{x^{2}}{2v}}$$
Assume that $m$ is even and $c_m$ is negative, so the definite integral of $e^{y(x)}$ is bounded. The coefficients $c_n$ are real. The result should be in terms of v and the $c_n$'s. The cases $m=0$ and $m=2$ are trivial.
My own, unsuccessful effort: The closest I've come to a solution is laboriously generating a Maclaurin series in $v$ for the $ln$ of the answer, using the following relations, which are true for any $y(x)$ due to properties of the Normal:
$$g(x,v)\equiv ln\left(\int_{-\infty}^{\infty}e^{y(z)}N(x-z,v)dz\right)$$
$$\frac{dg}{dv}=\frac{1}{2}\frac{d^{2}g}{dx^{2}}+\frac{1}{2}\left(\frac{dg}{dx}\right)^{2}$$
$$g(x,0)=y(x)$$
But this approach requires increasingly complex calculation of each term of the series, and I need an exact solution. Could this diff eq somehow be used to calculate an exact solution?