If $2x-y+c\log \left(\left(x-2y-4\right)\right)=k$ is the general solution of $\frac{dy}{dx}=\frac{2x-4y-5}{x-2y+2}$ then $c=$

$\frac{dy}{dx}=\frac{2x-4y-5}{x-2y+2}$

$=\frac{2\left(x-2y+2\right)-9}{x-2y+2}$

Let $x-2y+2=t\Rightarrow 1-2\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}-\frac{1}{2}\frac{dt}{dx}$

i.e. $\frac{1}{2}-\frac{1}{2}\frac{dt}{dx}=\frac{2t-9}{t}$

$\Rightarrow \frac{dt}{dx}=-2\left(\frac{2t-9}{t}-\frac{1}{2}\right)=\frac{18-4t}{t}+1$

$\Rightarrow \frac{dt}{dx}=\frac{18-3t}{t}\Rightarrow \frac{t}{18-3t}dt=dx$

$\Rightarrow -\frac{6-t-6}{6-t}dt=3dx$

$\Rightarrow -dt+\frac{6}{6-t}dt=3dx$

Integrating, we get

$-t-6\ln \left|6-t\right|=3x+C$

$-x+2y-2-6\ln \left|6-x+2y-2\right|=3x+C$

$-4x+2y-2-6\ln \left|4-x+2y\right|=C$

$2x-y+3\ln \left|x-2y-4\right|=k$

Hence, $c=3$