0
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I am a bit confused about how to solve an ODE of this type

$y'= \mid x \mid - \mid y \mid $ , $y(0)=0$

Do I need to divide the ODE into four phases :

  1. $x>0$ & $y>0$

  2. $x>0$ & $y<0$

  3. $x<0$ & $y>0$

  4. $x<0$ & $y<0$

or is there another approach?

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  • $\begingroup$ Note that the solution of this problem is unique, since the right hand side is Lipschitz continuous. So if you can somehow guess a solution and verify it, you are done. Why don't you try that ? :) $\endgroup$ – Hans Engler May 4 at 13:54
  • $\begingroup$ CAS says:$y(x)=\begin{cases} x-e^x+1 & x\leq 0 \\ x+e^{-x}-1 & x>0 \end{cases}$ $\endgroup$ – Mariusz Iwaniuk May 4 at 14:09
  • $\begingroup$ But those two equations (above) only answer the situation of (y>0). isn't that right? $\endgroup$ – baraah baryhe May 4 at 16:52

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