Try to solve the following differential equation: $y''-4y=2\tan2x$ I am trying to solve this equation:
$y''-4y=2\tan2x$
the Homogeneous part is:
$$y_h=c_1e^{2x}+c_2e^{-2x}$$
and I get according the formula:
$$C_1'e^{2x}+C_2'e^{-2x}=0$$
$$2C_1'e^{2x}-2C_2'e^{-2x}=2\tan2x$$
my questions is:


*

*if $y_h$ is right?

*how can I find $c_{1}$ , $c_{2}$ ?
thanks
 A: What you have done so far is correct. You should proceed as follows:
Write the last two equations as a system
$$\left(\begin{array}{cc}
e^{2x} & e^{-2x}
\\
e^{2x} & -e^{-2x}
\end{array}\right)
\cdot \left(\begin{array}{c} C_{1}^{\prime}
\\
C_{2}^{\prime}
\end{array}\right)=\left(\begin{array}{c}
0
\\
\tan(2x) \end{array}\right)$$
Invert the matrix on the LHS to get
$$\left( \begin{array}{cc}
C_{1}^{\prime}
\\
C_{2}^{\prime} \end{array}\right)=-\frac{1}{2}\left(\begin{array}{cc}
-e^{-2x} & -e^{-2x}
\\
-e^{2x} & e^{2x}\end{array} \right)\cdot \left( \begin{array}{c}
0
\\
\tan(2x) \end{array}\right)=-\frac{1}{2}\left(\begin{array}{c}
-e^{-2x}\tan(2x)
\\
e^{2x}\tan(2x) \end{array} \right)$$
This gives you a system of first order equations which you can solve by integrating. However, its not a particularly nice integral. 
A: ok let us solve homogeneous part, we have
$y''-4*y=0$
or characteristic equation is $k^2-4=0$ from where $k_1=2$ and $k_2=-2$ , so homegenous solution we have
$y_p(x)=c_1*e^{2*x}+c_2*e^{-2*x}$
$c_1$ and $c_2$ can be found using initial condition, like $y(0)=0$ and $y'(0)=-1$ for example, about $y''-4*y=2*\tan(2x)$
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx
use this method, but I have checked wolfram alpha and it has not solution in particular solution in real functions
