If we look at the differential equation, y' = 1 - 2y , we see that y' = 0 for y = 1/2 . This is what is sometimes called a "stationary solution" (also referred to as the "trivial" solution). If the initial value for y (for any time choice) were 1/2 , the value of y would remain at 1/2 forever. For any initial value y(0) < 1/2 , the function for y will have y' > 0 , with y' becoming smaller as y increases toward 1/2 . By the same token, for any initial value y(0) > 1/2 , y' < 0 , meaning that y will decline toward 1/2 , and at ever slower rates as y approaches that value.
This results in two cases for the IVP then:
for y(0) < 1/2 , the integration gives -0.5 ln ( 1 - 2y ) = x + C , which leads to an acceptable value for C because ( 1 - 2y ) is always positive;
but for y(0) > 1/2 , the integration gives -0.5 ln ( 2y - 1 ) = x + C , which makes the argument of the logarithm positive and maintains the values y will pass through within the domain of the function (and so will give a sensible value for C).
This is what is contained in the standard result for this problem, y = -0.5 ln | 1 - 2y | = x + C : the presence of the absolute value signals that there are two cases to consider (sadly, a matter often inadequately covered in introductory DE courses).