Let
\[f''++2g=4\]
(1)
\[f+g=0\]
(2)
We can obtain a single equation in \[f\]
by differentiating in (2) and substituting into (1).\[f''++2g=4\]
(1)
\[f'+g'=0\]
(3)From (3)
\[f'=-g'\]
then we can write (1) as \[f''-2f'=4\]
This last equation can bee solved using the integrating factor method.Multiply the equation throughout by
\[e^{\int{-2dx}}= e^{-2x}\]
.We obtain
\[f' e^{-2x}-2f e^{-2x}=4e^{-2x} \]
.We can rewrite the equation as
\[(fe^{-2x})'=4e^{-2x}\]
.Integrate both sides.
\[fe^{-2x}=-2e^{{-2x}}+ C\rightarrow f=-2+Ce^{2x}\]
then from (2) \[g=2-Ce^{2x}\]
.