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.