Solve $f''(x)+4f'(-x)=e^{2x}$ 
Solve $f''(x)+4f'(-x)=e^{2x}$

Separating the function into odd and even parts yields two equations which can be solved separately.
But then, how to get the solutions of the equations from the two equations?
The even equation $f_e''(x)-4f_e'(x)=e^{2x}$ has no even solutions. Does that mean the above equation also has no solutions?
 A: So you got to
$$
(f_e''(x)+f_o''(x))+4(-f_e'(x)+f_o'(x))=e^{2x}
$$
Separating again into even and odd parts gives (the derivative of an even function is odd, and vv.)
$$
f_e''(x)+4f_o'(x)=\cosh(2x),\\
f_o''(x)-4f_e'(x)=\sinh(2x).
$$
Now one could set $g(x)=f_e'+if_o'$, $g(-x)=-\overline{g(x)}$, so that then
$$
g'(x)-4ig(x)=\cosh(2x)+i\sinh(2x)\implies g(x)=A\cosh(2x)+B\sinh(2x)+Ce^{4ix}.
$$
Because of 
$$
-\overline{g(-x)}=-\bar A\cosh(2x)+\bar B\sinh(2x)-\bar C e^{4ix}
$$ 
we get $C=ic$, $A=ia$ imaginary, $B=b$ real. Now insert and solve
$$
2ia\sinh(2x)+2b\cosh(2x)+4a\cosh(2x)-4ib\sinh(2x)=\cosh(2x)+i\sinh(2x)\\
2a-4b=1\\
2b+4a=1\\
10a=3\\
10b=-1
$$
so that finally
$$
f_e'=-0.1\sinh(2x)-c\sin(4x)\implies f_e=-0.05\cosh(2x)+\tfrac{c}4\cos(4x)+d\\
f_o'=0.3\cosh(2x)+c\cos(4x)\implies f_o=0.15\sinh(2x)+\tfrac{c}4\sin(4x)
$$
A: Take the derivative of the equation
$$
f'''(x)-4f''(-x)=2e^{2x},
\\
f''(-x)+4f'(x)=e^{-2x}
\\
\implies f'''(x)+16f'(x)=2e^{2x}+4e^{-2x},
$$
which now is a normal scalar linear differential equation. It has solutions of the form
$$
f(x)=A\cos(4x)+B\sin(4x)+C+De^{2x}+Ee^{-2x}
$$
Inserting this into the original equation gives
\begin{align}
&-16(A\cos(4x)+B\sin(4x))+4(De^{2x}+Ee^{-2x})
\\
&+16(A\sin(4x)+B\cos(4x))+8(De^{-2x}-Ee^{2x})
\\\hline
&=e^{2x}
\end{align}
so that $A=B$ and $E=-2D$, $1=4D-8E=20D$. In total
$$
f(x)=A\bigl(\cos(4x)+\sin(4x)\bigr)+C+\frac1{20}(e^{2x}-2e^{-2x})
$$
