how to solve this BVP $-\dfrac{d^2u}{dx^2} +2u = 1$ $u(0)=u(1)=1$ using a Fourier sine series?

Question

I know how to solve a BVP in the form of $$ -\dfrac{d^2u}{dx^2} = 1 $$ using a Fourier sine series but i don't know how to go about solving $$-\dfrac{d^2u}{dx^2} +2u = 1 $$ My guess is that i have to use the same formulas but the function i plug in has to be different, it can't be 1 or i'd get the same fouier series as $$-\dfrac{d^2u}{dx^2} = 1$$ Thanks.

Answer 1

Set $$ v = 2u -1 \implies v'' = 2u'' $$ in the equation $$ -\frac{1}{2}v'' +v =0 \implies v'' =2v $$ which solution is quite easy