#### crashtest84

##### New member
I don't understand how they are justifying the replacement that is being made in this example. It makes sense but I can't quite grasp the logic that's being used here so I don't think I'd be able to implement this method on other problems. Any help would be appreciated!

#### crashtest84

##### New member
Replacing absolute value

This sort of makes sense when I see what they've done but I don't understand the logic well enough to use this method myself. It seems a little arbitrary. Any help would be appreciated!

#### HallsofIvy

##### Elite Member
I have no idea what you are asking! You post what appears to give an "alternative method" to solving a problem but you don't post the problem itself.

#### HallsofIvy

##### Elite Member
?? Is this the same problem as in your thread "Replacing absolute value"? Again, you show just part of the information. Apparently this involves a first order differential equation with a given initial condition, apparently of the form y(-1)= some given number. If that is correct then the "logic" is that you have found the general solution to the differential equation to be of the form y(x)= F(x, C) where C is a "constant of integration", the undetermined constant that occurs when integrating (because the derivative of any constant is 0). In order to determine what C must be we must get an equation involving only C. To do that we replace x and y with the values given in the initial condition. If that initial condition is of the form y(-1)= A then replace x by -1 and y by A to get an equation in C only.

I can't say more than that without knowing what the differential equation (or at least the general solution) is and what the initial condition is.

#### Dr.Peterson

##### Elite Member
This sort of makes sense when I see what they've done but I don't understand the logic well enough to use this method myself. It seems a little arbitrary. Any help would be appreciated!

View attachment 10640

Here we have replaced |y - 1| by 1 - y and |x + 3| by x + 3, since we are interested in x and y near the initial values x = -1, y = 0 (for such values, y - 1 < 0 and x + 3 > 0).

then you are given an explanation, but perhaps it needs a little more.

Apparently in the problem you didn't show, you only need to find a solution that includes (-1, 0), so you can restrict your attention to points near there.

If x is near -1, then x + 3 is near 2, and therefore is positive (taking "near" to be "no farther away than 2 units"). The absolute value of a positive number is the number itself, so |x + 3| = x + 3 in this case.

If y is near 0, then y - 1 is near -1, and therefore is negative (if y is within 1 of 0). The absolute value of a negative number is its negative, so in this case |y - 1| = -(y - 1) = 1 - y.

That is what they are saying.

Is that what you were asking? (It helps to both state the context, and be specific in what you ask.)