Is there a faster way (I've found a good one but I want to be sure) - interesting calculus question I quite liked this question but I did something I don't normally do to make it... faster to write and thus do, I'm wondering however if there's a faster way.
Given:
$$y(x)=e^{-x}\int_0^{e^{2x}}{\left(\frac{1}{t}\int_0^t\left(\frac{g(s)}{\sqrt{s}}\right)ds\right)}dt$$
RELAX you need not work this out (I panicked a bit when I first saw it I must confess - also bracket height not changing in LaTeX? I expected the outer ( ) to be taller and contain their contents..)
First:
Find $y'$ and $y''$
Then:
Find $a,b,c\in\mathbb{R}$ such that $ay''+by'+cy=4g(e^{2x})$
To make it easier I let 
$$\alpha(x)=e^{2x}$$
$$F(\beta)=\int^\beta_0\left(\frac{1}{t}\int^t_0\left(\frac{g(s)}{\sqrt{s}}\right)ds\right)dt$$
$$\Phi(\gamma)=\int^\gamma_0\left(\frac{g(s)}{\sqrt(s)}\right)ds$$
Then one may write:
$$y=e^{-x}F(\alpha(x))=e^{-x}F(\alpha)$$
Now using the fundamental theorem of calculus things like: (where the lower case letter is the derivative of the upper, that is f(x) is d(F(x))/dx)
$$f(\beta)=\frac{1}{\beta}\Phi(\beta)$$
happen, the chain rule can be used to state:
$$\frac{d}{dx}[F(\alpha)]=f(\alpha).\alpha'$$
(well that is basically the chain rule)
So the question becomes manageable quickly, you do the differentiation without error then you compare coefficients to find a,b and c to be 2,4 and 2 respectively. 
I'd like the answer confirmed, and to identify any other methods that would work just as well.
Here are the values:
$y=e^{-x}F(\alpha)$
$y'=-e^{-x}F(\alpha)+e^{-x}\Phi(\alpha)$
$y''=e^{-x}F(\alpha)-2e^{-x}\Phi(\alpha)+2e^x\phi(\alpha)$
BUT:
$\phi(z)=\frac{g(z)}{\sqrt{z}}$
Thus:
$y''=e^{-x}F(\alpha)-2e^{-x}\Phi(\alpha)+2g(\alpha)$
Addendum:
I actually slipped up early into the problem with my expression for $y'$ that was then carried forward, the method is sound but it lead to me getting double the answer I ought to have gotten for the final values, see Turnococ's answer.
 A: It is good that you split your function into parts.
\begin{align}
y(x) & = e^{-x} F(e^{2x}) \\
F'(x) & = f(x) = \frac 1x\Phi(x)\\
\Phi'(x) & = \phi(x) = \frac{g(x)}{\sqrt x}
\end{align}
Using these equations, we can compute
\begin{align}
y'(x)
& = -e^{-x}F(e^{2x}) + 2e^{x}f(e^{2x})\\
& = -y(x) + 2e^{x}f(e^{2x}) \\
& = -y(x) + 2e^{-x}\Phi(e^{2x}) \\
y''(x) & = -\left(-y(x) + 2e^{-x}\Phi(e^{2x})\right)
- 2e^{-x}\Phi(e^{2x})
+ 4e^{x}\phi(e^{2x})\\
& = y(x) - 4e^{-x}\Phi(e^{2x})
+ 4e^{x}\phi(e^{2x}) \\
& = y(x) - 4e^{-x}\Phi(e^{2x})
+ 4g(e^{2x})
\end{align}
For ease of notation, let $u(x) = y(x)$, $v(x) = e^{-x}\Phi(e^{2x})$ and $w(x) = g(e^{2x})$.
Then, we can express $y, y', y''$ using $\left\{u, v, w\right\}$ as the basis:
\begin{align}
y & =
\begin{pmatrix}
u & v & w
\end{pmatrix}
\begin{pmatrix}
1 \\ 0 \\ 0
\end{pmatrix}\\
y' & =
\begin{pmatrix}
u & v & w
\end{pmatrix}
\begin{pmatrix}
-1 \\ 2 \\ 0
\end{pmatrix}\\
y'' & =
\begin{pmatrix}
u & v & w
\end{pmatrix}
\begin{pmatrix}
1 \\ -4 \\ 4
\end{pmatrix}\\
\end{align}
Solving for $a, b, c$ such that $ay'' + by' + cy = 4g(e^{2x})$ becomes solving the linear system
\begin{align}
\begin{pmatrix}
1 & -1 & 1\\
0 & 2 & -4\\
0 & 0 & 4
\end{pmatrix}
\begin{pmatrix}
a \\ b \\ c
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 0 \\ 4
\end{pmatrix}
\end{align}
The solution is $(a, b, c) = (1, 2, 1)$.
Note that the solution is unique only if $u, v, w$ are linearly independent. I believe it is possible to derive, from computing the Wronskian of $u, v, w$, a condition (or conditions) under which $u, v, w$ will be independent, but I honestly do not want to go there.
