Recurrence Relation

Set $f(x) = \sum_{n = 0}^\infty a_n z^n$, then $$f''(z)\cdot z +2 f'(z) = \frac{d^2}{dz^2} (z\cdot f(z)) = \frac{d^2}{dz^2}(a_0+a_1z+\cdots) = 2a_1+6a_2z+\cdots $$ $$= 2a_1+\sum_{k = 2}^\infty k(k+1)a_k z^{k-1} = \sum_{k = 2}^\infty 2(\lambda k -1 )a_{k-1}z^{k-1}+\sum_{k = 2}^\infty(a-\lambda^2)a_{k-2}z^{k-1} $$ $$= 2a_1+2\lambda \sum_{k = 2}^\infty k a_{k-1}z^{k-1} - 2\sum_{k = 2}^\infty a_{k-1}z^{k-1}+(a-\lambda^2)\sum_{k = 2}^\infty a_{k-2}z^{k-1}$$ $$= 2a_1+2\lambda\frac{d}{dz}(f(z)\cdot z) - 2\lambda a_0- 2f(z)+2a_0 + (a-\lambda^2)f(z)z$$

Solve this second order differential equation and find a solution holomorphic at 0. After you obtain solution $f(z)$ we have $a_n = \frac{f^{n}(0)}{n!}$. But this seems to be too complicated.