Thanks for the explanation! Hmmm... I still don't really get the second question; you don't usually put modulus signs when you simply rearrange the equation :/ Or do you have to in this case since there is a "c"?
For your first question with dy/dx = y
2, I think it should be x≠2. If you graph the function, the gradient is always positive (and y
2) so I don't think there's any real need to only take x<2 unless there was some other condition such as it having to be continuous
For the second question, you shouldn't need to put in modulus signs, from previous working x-c is negative so c-x should be positive. Modulus signs should usually only appear in integration (for 1/x or similar). I think it's probably an error, as I don't believe c-x < 0 would work.
Also when I graphed the solution they gave, half the graph had a negative gradient which is not possible as e^anything > 0. Without the modulus signs it's always positive though.