Variation of parameters (issue with the constants) I have to solve this ODE using the variation of parameters method:
$$4y''+y=\frac{2}{\cos \left( \frac{x}{2} \right)}$$
Solving the homogeneous problem yields
$$y_h(x)=
c_1 \cos \left( \frac{x}{2} \right)+
c_2 \sin \left( \frac{x}{2} \right)$$
Now, to solve the variation of parameters problem, you have to solve
$$
\begin{bmatrix}
  \cos \left( \frac{x}{2} \right) &
  \sin \left( \frac{x}{2} \right) \\
 -\frac{1}{2} \sin \left( \frac{x}{2} \right) &
  \frac{1}{2} \cos \left( \frac{x}{2} \right)
\end{bmatrix}
\begin{bmatrix} c_1' \\ c_2' \end{bmatrix}=
\begin{bmatrix}
  0 \\
  2\sec \left( \frac{x}{2} \right)
\end{bmatrix}$$
Solving the first equation $
c_1'\cos \left( \frac{x}{2} \right)+
c_2'\sin \left( \frac{x}{2} \right)=0$. This gives 
$c_1'=-c_2'\tan \left( \frac{x}{2} \right)$
Solving the second equation gives $
\frac{c_2'}{2} \cos \left( \frac{x}{2} \right)-
\frac{c_1'}{2} \sin \left( \frac{x}{2} \right)=
\frac{2}{\cos \left( \frac{x}{2} \right)}$.
Substituting gives $
\frac{c_2'}{2} \cos \left( \frac{x}{2} \right)+
\frac{c_2'}{2}
\frac{\sin^2 \left( \frac{x}{2} \right)}
     {\cos \left( \frac{x}{2} \right)}=
\frac{2}{\cos \left( \frac{x}{2} \right)}$.
Solving for $c_2'$ gives $\frac{c_2'}{2}=2$. 
Finally, $c_2'=4$ and $c_2=4x$. This means that $
c_1'=-4\tan \left( \frac{x}{2} \right)$ and $
c_1=8\ln \left( \cos \left( \frac{x}{2} \right) \right)$.
However, according to Wolfram, $c_1=2\ln \left( \cos\left( \frac{x}{2} \right)\right)$ and $c_2=x$.
Indeed, I tried solving with my values for $c_1$ and $c_2$ and it doesn't work. What did I do wrong?

Try with the equation writen in this form. The method needs the coefficient to be 1 for the highest derivative (standard form).
$$y''+\frac{y}{4}=\frac{1}{2\cos \left( \frac{x}{2} \right)}$$
With it, we get $c_2'/2=1/2$ and $c_1'=-\tan \left( \frac{x}{2} \right)$. With them, the expected solution follows.
