Pratically we have a classical first order linear DE...

$\displaystyle g^{'} (x) = -g(x) + 3 x^{2} +4x +2$ (1)

... of which we are searching a solution in some sense 'quadratic'. If we proceed in 'standard fashion' we find that...

$\displaystyle g(x)= c\cdot e^{-x} + 3 x^{2} -2 x + 4 $ (2)

... where c is an arbitrary constant. Now may be that for 'quadratic' we intend that it must be $\displaystyle c=0$... or may be some else

...

Merry Christmas from Italy $\displaystyle \chi$ $\displaystyle \sigma$