Differential Equation Problem

You don't know what to add to 2 to get 1? Is that your problem? Draw a number but do Not put any numbers on it but you can put some slashes on the number line. Now put your pen on one of those slash marks and go two places to the right since you want to add two to this unknown number. Than after going two places, put a 1 by that position (since the unknown number + 2 = 1). Now count back two places from one and you'll know this unknown number. Can e^(1+c) equal that number. Or is that where you are stuck?

Can you please explain what rule you used to go from -ln| 2-y| = x to e^(x+c)?
 
First, where you have "-ln|2- y|= x" you should have "-ln|2- y|= x+ c". Also the exponential of -ln|2- y| is NOT -(2- y)= -2+ y. Rather, -ln|2- y|= ln|(2- y)^{-1}| though I would be inclined to rewrite "-ln|2-y|= x+ c" as "ln|2- y|= -x- c" and then take the exponential: |2- y|= e^{-x- c}.

Now, I am sure that you understand that e^{-x- c}= e^{-x}e^{-c}. And since "c" is just an "undetermined constant" so is e^{-c} so we can call it, say, c'. Also, although e to any power, positive or negative, is always positive, we can remove the absolute value from |2- y| by allowing c' to be negative. That is 2- y= c'e^{-x} where c' can be any number. Then y= 2- c'e^{-x}.
 
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