First ODE of an absolute value

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SUMMARY

The discussion focuses on solving the first-order ordinary differential equation (ODE) y' - y = |x - 1|. The solution involves breaking the problem into two cases based on the value of x: for x ≥ 1, the equation simplifies to y' - y = x - 1, and for x < 1, it becomes y' - y = 1 - x. Each case is integrated separately to obtain general solutions valid for their respective intervals.

PREREQUISITES
  • Understanding of first-order ordinary differential equations (ODEs)
  • Knowledge of integration techniques
  • Familiarity with absolute value functions
  • Basic differential equations terminology
NEXT STEPS
  • Study integration methods for first-order ODEs
  • Learn about piecewise functions and their applications in differential equations
  • Explore the concept of continuity and differentiability in relation to ODEs
  • Practice solving more complex ODEs involving absolute values
USEFUL FOR

Students preparing for exams in calculus or differential equations, educators teaching ODEs, and anyone seeking to strengthen their understanding of solving equations involving absolute values.

sydneyw
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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.
 
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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.
 
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 x\ge 1, 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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