2006 AB Calculus Exam Section II 5b solution

During the last 3 days, students, and teachers were talking about the solution to free-response question 5b online or in the classroom. Different answers have been presented. which one is the right answer? If you still trust me, read my solution.

Solution to free-response question 5b:

Given dy/dx=(1+y)/x where x ≠ 0
Solve the differential equation. We get,
ln[abs(1+y)] = ln[abs(x)] + C
e^abs(1+y) = e^( ln[abs(x)] + C)
abs(1+y)=e^C* e^ln[abs(x)]
1+y=±e^C * abs(x)
1+y=k *abs(x) where k = ±e^C, and k is a constant.
y=k abs(x) - 1 where x ≠ 0 ( Note: x ≠0 is given)
Using the initial condition f(-1)=1, we get k=2.
So the solution is:
y=2 abs(x) - 1 where x ≠ 0 or the domain is all real numbers except for x=0.
Since the solution is a piecewise function, it can also be expressed as
y= -2x-1 when x<0
y= 2x-1 when x>0

Check the original equation with each piece.
When x<0, y=-2x-1
dy/dx=(1+y)/x
(-2x-1)' = ( 1 + (-2x-1 ))/x
-2 = -2 (True)

When x>0, y=2x-1
dy/dx=(1+y)/x
(2x-1)' = ( 1 + (2x-1 ))/x
2 = 2 (True)

Therefore, The solution is: y=2x - 1 where x ≠ 0.