Question about replaced values

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.)