Do the Solutions to Integrals Depend on Which Side has the Constant?

Question

I am trying to solve the following integral problem: $\int$$\frac{dx}{x}$ $=$ $\int[\frac{1}{y-1}-\frac1y]$$dy$. Applying the integral should yield: $\ln\lvert x\rvert$ $=$ $\ln\lvert y-1\rvert$ - $\ln\lvert y\rvert$ $+$ $C$. I then raised the right hand side to $e$ so I could obtain $\lvert x \rvert$. It looks like this: $\lvert x\rvert$ $=$ $e^{{(\ln\lvert y-1\rvert}/\ln\lvert y\rvert) +C} $ After simplifying the $ln$ terms raised to a power and getting rid of the absolute value signs I get: $C\frac{y-1}{y} = x$ In solving for $y$ I get $y$ = $\frac{C}{C-x}$. The correct solution is $\frac{1}{1-Cx}$. I checked to see if my solution is equivalent, but it is not. In the work of this problem it appears that the solver put $C$ on the side of $\ln\lvert x\rvert$, -- the opposite side of my constant. Why do these two ways yield different results and why is my way of solving incorrect.

Answer 1

It the same, just divide numerator and denominator by $C$: $$ y=\frac{C}{C-x} = \frac{C/C}{C/C - x/C} = \frac{1}{1-x/C} $$

Since $C$ is just an integration constant, $K = 1/C$ would be an integration constant as well, so $$ y=\frac{1}{1-Kx} $$ would be a equivalent form.

Answer 2

Obviously not,

$$f(x)+c=g(y)\text{ and }f(x)=g(y)+c$$

are identical, but for the sign of $c$, which is arbitrary.