## Simultaneous Differential Equations

==],k]]n happens that two differential equations are expressed in terms of the same functions. I may be possible to find a single equation in terms of a single function, which may then be solved to find a solution.
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)
$g=2-Ce^{2x}$
.