Formally write $$u(x,\lambda) = \sum_{j=0}^\infty u_j(x) (\lambda - 1)^j$$ The coefficients of each power of $\lambda-1$ in the differential equation give you a sequence of d.e.'s for the $u_j$'s: $u_0'' + u_0' - i V u_0$ (so that $u(x,1) = u_0(x)$ is a solution of your d.e. for $\lambda = 1$), and $u_j'' + u_j' - i V u_j = i V u_{j-1}$ for $j \ge 1$. Thus $u_j$ is a solution of the inhomogeneous linear equation corresponding to your homogeneous equation, with forcing term depending on $u_{j-1}$.
Of course, to get a particular solution you need to specify initial or boundary conditions, and you have not done so. Convergence or divergence of the series may depend on those conditions.