#### crashtest84

##### New member

- Joined
- Dec 10, 2018

- Messages
- 4

- Thread starter crashtest84
- Start date

- Joined
- Dec 10, 2018

- Messages
- 4

- Joined
- Dec 10, 2018

- Messages
- 4

- Joined
- Jan 27, 2012

- Messages
- 4,934

- Joined
- Jan 27, 2012

- Messages
- 4,934

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.

- Joined
- Nov 12, 2017

- Messages
- 3,590

If you are specifically asking only about this line,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.)

- Joined
- Dec 10, 2018

- Messages
- 4

That's the part that was confusing. I couldn't quite make the logically connection you've spelled out and I get it now. Thanks for the help!If you are specifically asking only about this line,

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.

- Joined
- Dec 10, 2018

- Messages
- 4

Double-post, sorry about that. The explanation in the other thread cleared up my confusion.