Hello Brilliant I'm back!

First of all let's recall what the Wronskian of two function is: $W(f,g)=f g'-gf'$ . So we have a system of three relations:

$\begin{cases} fg'+gf'=\left(5x^2+\frac{107}{6}x+\frac{47}{3}\right) e^{5x} \\ fg'-gf'=\left(-x^2-\frac{19}{6}x-\frac{7}{3}\right) e^{5x} \\ \frac{f'}{g'}=\frac{3}{2}e^x \end{cases}$

Subtracting the second equation to the first we get:

$2gf'=\left(6x^2+21x+18\right) e^{5x}$

and using the third eqn:

$2g\cdot \frac{3}{2}e^xg'=3\left(2x^2+7x+6\right) e^{5x} \Rightarrow gg'=\left(2x^2+7x+6\right) e^{4x}$

Integrating both parts: $\frac{g^2(x)}{2}=\frac{1}{8}\left(2x+3\right)^2 e^{4x}\Rightarrow g^2(x)=\frac{1}{4}\left(2x+3\right)^2 e^{4x}\Rightarrow g(x)=\frac{1}{2}\left(2x+3\right) e^{2x}$

Note: we can discard the negative solution because $(f+g)(1)$ in the end is requested to be positive.

Solving the third eqn for $f(x)$ : $f'(x)=3(x+2)e^{3x}\Rightarrow f(x)=\frac{1}{3}(3x+5)e^{3x}$ .

It's easy to check that this definition for $f(x)$ and $g(x)$ verifies the three eqns of the system.

Now we only need to evaluate the sum $f(x)+g(x)$ at $x=1$ :

$f(1)+g(1)=\frac{5 e^2}{2}+\frac{8 e^3}{3}$ ,

so

$A=5, B=2, C=2, D=8, F=3, G=3$ and $A+B+C+D+F+G=23$