Differential equations: solving separable equation Solve the separable equation $y' = (x-8)e^{-2y}$ satisfying the initial condition $y(8)=\ln(8)$. I can not figure this out I am not sure what I am doing wrong.

We have $$\frac{dy}{dx}=(x-8)e^{-2y}\implies e^{2y}dy=(x-8)dx$$
Integrating either sides 
$$\frac{e^{2y}}2=\frac{x^2}2-8x+C\implies e^{2y}=x^2-16x+2C$$ where $C$ is an arbitrary constant
Putting $x=8, y=\ln8, e^{2\ln 8}=8^2-16\cdot8+2C$
$2C-64=(e^{\ln 8})^2=8^2\implies 2C=64+64$
