A student attempted to solve the differential equation $$\displaystyle \frac{dy}{dx} = \frac{1-y}{x}$$ with the initial condition that $$y=0$$ when $$x=1$$. The student’s work is shown below. In which step does an error first appear?

Step 1: $$\displaystyle \int \frac{1}{1-y} dy = \int \frac{1}{x} dx$$

Step 2: $$-\ln |1-y| = \ln |x| + C$$

Step 3: $$\ln |1-y| = -\ln |x| - C$$

Step 4: $$\ln |1-y| = \ln \left|\frac{1}{x}\right| - C$$

Step 5: $$|1-y| = \frac{K}{x}$$ where $$K = e^{-C}$$

Step 6: Using the initial condition: $$|1-0| = \frac{K}{1}$$, so $$K = 1$$

Step 7: $$|1-y| = \frac{1}{x}$$

Step 8: $$1-y = \pm\frac{1}{x}$$

Step 9: Since $$y = 0$$ when $$x = 1$$, we have $$1-y = \frac{1}{x}$$

Step 10: $$y = 1 + \frac{1}{x}$$