Solving $u_{yy} + (2-x)u_y - 2xu = 1$

Question

I want to solve the pde $$ u_{yy} + (2-x)u_y - 2xu = 1 $$ so if I treat $x$ in the coefficients as arbitrary but fixed it is equivalent to solving the ode $$ y'' + (2-x) y' - 2x y = 1. $$ For the homogenous part $y'' + (2-x) y' - 2x y = 0$ I have to solve the characteristic equation $$ \lambda^2 + (2-x)\lambda - 2x = 0 $$ which has solutions: $$ \lambda_{1/2} = \frac{2-x}{2} \pm \sqrt{ \frac{(2-x)^2}{4} + 2x } = \frac{2-x}{2} \pm \sqrt{ \frac{x^2 + 4x + 4}{4} } = \frac{2-x}{2} \pm \frac{x+2}{2} $$ But I have difficulty finding a particular solution, any hints?