I need to find the orthogonal trajectory of:
$$x^2+y^2=Cx$$
Here is my attempt:
$ \begin{align} 2x+2yy'&=c\\ x+yy'&=c\\ yy'&=c-x\\ y'&=\frac {c-x} y \text{ <- DE of original}\\ y'&=-\frac y {c-x} \text{ <- DE of ot}\\ \frac {dy}{dx}&=-\frac y {c-x} \\ \int \frac {dy}{y}&=-\int \frac {dx} {c-x} \\ -\ln|y|&=-\ln|c-x|+\ln|c| \\ \ln|y|&=\ln|c-x|-\ln|c| \\ |y|&=|c-x|-|c| \\ y&=K-x-c \\ \end{align} $
I attached a picture with my work and was hoping for someone to help me check whether it is right.
Given equation: $$\ x^2 + y^2 = cx$$ Differentiating it wrt$\ x$ we get:$$\ 2x + 2y(\frac{dy}{dx}) = c$$Now replace $\ \frac{dy}{dx}$ by$\ \frac{-dx}{dy}$ to get differential equation of orthogonal trajectory which is:$$\ \frac{-2dx}{c-2x} = \frac{dy}{y}$$Now integrate both sides:$$\int \frac{-2dx}{c-2x} = \int \frac{dy}{y}$$$$\ ln|c-2x| = lny + lnk$$$$\ ln|c-2x| = ln|ky|$$$$\ y = \frac{c-2x}{k}$$ where$\ k$ is any constant.