so $C=-ln|50|$

and $-ln|50-y|=kx-ln|50|$

$ln|50-y|-ln|50|=-kx$

so $ln\left|1-\frac{y}{50}\right|=-kx$

$1-\frac{y}{50}=e^{-kx}$

so $y=50(1-e^{-kx})$