How to solve this differential equation I've got an equation
$$e^x(y^2-4y' -4y)=4y'$$
after some transformations, i recieve
$$y= \frac{-4}{e^x+C_2(x)}$$
$$y'=\frac{4(e^x+C'_2(x))}{(e^x+C_2(x))^2}$$
but when i add these equations to the first one $C_2(x)$ sill remains in equation.. Can somebody sole it?
 A: As per the comment discussion on the question, $C_2(x)$ should really be $C_2$ since $C_2$ is just a constant and does not depend on $x$.
You found $y(x) = \dfrac{-4}{e^x + C_2}$.
So then $y'(x) = \dfrac{(e^x + C_2)(0) - (-4)(e^x)}{(e^x + C_2)^2} = \dfrac{4e^x}{(e^x+C_2)^2}$
When you plug this into the LHS of the original equation, you get:
\begin{align*}
  e^x(y^2 - 4y' - 4y) &= e^x \left(\frac{16}{(e^x+C_2)^2} - \frac{16e^x}{(e^x+C_2)^2} + \frac{16}{e^x+C_2}\right)\\[0.3cm]
  &= e^x \left(\frac{16}{(e^x+C_2)^2} - \frac{16e^x}{(e^x+C_2)^2} + \frac{16(e^x+C_2)}{(e^x+C_2)^2}\right)\\[0.3cm]
  &= e^x \left(\frac{16}{(e^x+C_2)^2} - \frac{16e^x}{(e^x+C_2)^2} + \frac{16e^x}{(e^x+C_2)^2} + \frac{16C_2}{(e^x+C_2)^2}\right)\\[0.3cm]
  &= e^x \left(\frac{16}{(e^x+C_2)^2} + \frac{16C_2}{(e^x+C_2)^2}\right)\\[0.3cm]
  &= \frac{16e^x(1+C_2)}{(e^x+C_2)^2}
\end{align*}
But in the RHS, you get:
$$ 4y' = \frac{16e^x}{(e^x+C_2)^2}
$$
So very close, but these are not quite the same.  So your $y(x)$ isn't correct.
