Question

After solving the differential equation xy'=y-1, I end up with y-1 = x + C. How does this become y = Cx + 1?

Answer 1

Separating variables
\[\int\limits \frac{dy}{y-1} = \int\limits \frac{dx}{x}\]

Answer 2

Now integrating both sides we have

ln(y-1) = ln(x) + C = ln(Ax), where A = e^C

hence y -1 = Ax

that is why the constant is the coefficient of x rather than being a floating constant.