I have a recursive equation which turns into this linear functional DE. $\Big[1 + a - az\Big]G(z) + \Big[bz - bz^2\Big]G^{\prime}(z) = \Big[a + (1-a)z\Big]G(1-c+cz), G(1) = 1$
where I assume $G(z)=\sum_{k \geq 0} a_k z^k $ and $a, b, c$ are some constants.
I am trying to solve this DE analytically but it seems to be hard. Is is possible to solve this special kind of functional DE or I should use numerical methods? Is there any available tool to solve these sort of linear functional DE numerically?
The fixed point of the transformation $z \mapsto 1-c+cz$ is $1$, so it's most natural to look at series around $z=1$. With $z = 1 + t$ and $G(z) = g(t)$, the equation becomes $$ (1-at) g(t) -bt(1+t) g'(t) = (1+(1-a)t) g(ct) $$ If we write $g(t) = \sum_{k=0}^\infty a_k t^k$ and expand this out, from the coefficient of $t^k$ we get $$ a_k - a a_{k-1} - b k a_k - b (k-1) a_{k-1} = c^k a_k + (1-a) c^{k-1} a_{k-1} $$ so that $$ a_{k} = \frac{(a-1) c^{k-1} + (1-k) b - a}{c^k + b k - 1} a_{k-1} $$ with $a_0 = 1$. Note that you will run into trouble if the denominator is ever $0$.
EDIT: Note that the radius of convergence of the series is the limit, if it exists, of $|a_{k-1}/a_k|$ as $k \to \infty$. This will be $|c/(a-1)|$ if $|c| > 1$ and $a \ne 1$, or $1$ if $|c| \le 1$ and $b \ne 0$. That doesn't necessarily mean the solution doesn't exist outside the radius of convergence, just that you can't use partial sums of the series to approximate it there.