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First ODE of an absolute value

  1. May 3, 2012 #1
    so I understand the basic premise of differentiating a first ODE, or I thought I did. I have the equation y'-y=abs(x-1). I have no idea of how to go about this. Can someone walk me through how to do this? I'm attempting to study for a test and this is one of the practice questions he gave us so I feel as though I'm in some serious trouble if I don't learn how to do this! Thank you much.
     
  2. jcsd
  3. May 3, 2012 #2

    LCKurtz

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    Can you solve ##y'-y=x-1##? Can you solve ##y'-y=1-x##? These are the two cases you have depending on whether ##x>1## or ##x<1##. Solve them separately.
     
  4. May 3, 2012 #3

    HallsofIvy

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    First, you don't want to "differentiate" the ODE, you want to integrate it.

    And the simplest way to handle the absolute value is to use the definition. If [itex]x\ge 1[/itex], x- 1 is non-negative so |x- 1|= x- 1 and your differential equation becomes dy/dx- y= x- 1.
    If x< 1, x- 1 is negative so|x- 1|= -(x- 1)= 1- x and your differential equation becomes dy/dx= 1- x.

    Integrate those to get two general solutions, one valid for x> 1, the other valid for x< 1.
     
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