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How did they get |x|=±x? How did they know y=0 is also a solution?

  1. Mar 13, 2013 #1
    Consider example 3 in the attachment. Better quality version: http://postimage.org/image/fjz61dbyr/

    How did they get ##|x|=±x##? If I remember correctly, I was told on physics forums that ##|x|≠±x##. The curious thing about this one is that it doesn't even involve square roots!

    How did they know y=0 is also a solution?
    Last edited by a moderator: Mar 14, 2013
  2. jcsd
  3. Mar 13, 2013 #2


    Staff: Mentor

    They didn't.

    They have |y| = eCe(1/3)x3 = Ae(1/3)x3, where A is a positive constant.

    This means that y = Ae(1/3)x3 or y = -Ae(1/3)x3, with A again being positive.
    By noticing that y = 0 satisfies the differential equation.
  4. Mar 13, 2013 #3


    User Avatar
    Homework Helper

    If we are talking about functions and graphs, then the graph of y = |x| isn't the same as the graph of y = ±x. But this is something different. It's essentially an absolute value equation, where if [itex]| x | = b[/itex] then [itex]x = \pm b[/itex].
  5. Mar 13, 2013 #4
    I don't understand how you went from:


    without the use that ##|x|=±x## where x is anything (not the same x in the example).

    $$y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } },\quad A>0\\ Set\quad y=0\\ 0=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }$$

    But there is no solution?
  6. Mar 13, 2013 #5


    Staff: Mentor

    Let's make it simple. Assuming that B is a positive number, the equation |y| = B is equivalent to y = B or y = -B.

    That's all they did.

    The original differential equation is y' = x2y.
    Notice that if y = 0, then y' = 0, so substituting this function and its derivative into the differential equation yields a true statement. That means that y = 0 is a solution of the DE.
  7. Mar 13, 2013 #6
    But that's not the equation. The equation is the first one: dy/dx = x^2y, y=0 satisfy the equation.
  8. Mar 13, 2013 #7
    So in other words, you're using the fact that ##|y|=B## is the same as writing ##y=±B##, right?

    Like this?

    $$\frac { dy }{ dx } ={ x }^{ 2 }y\\ Set\quad y=0\\ \frac { dy }{ dx } =0\\ ∴\quad when\quad y=0,\quad y'=0$$

    And then we substitute ##y=0## and ##y'=0## back into the differential equation to get a true statement? But we got those values from the DE, so obviously it will be true... Isn't this circular logic?
  9. Mar 13, 2013 #8
    What does it mean for a function to be a solution to a differential equation?
  10. Mar 13, 2013 #9
    Assuming that a differential equation works like a regular equation, a solution is one that consists of 3 values: one for x, one for y, and one for y' that satisfy the equation when plugged in.
  11. Mar 13, 2013 #10
    Okay, so I get that y=0 satisfies the DE.

    But why is it implied that just because it satisfies the DE, that the solution also works for y=0? I thought the solution was restricted to not work for y=0?
  12. Mar 14, 2013 #11


    Staff: Mentor

    Huh? The last part makes no sense. y = 0 is a solution of the DE, because it satisfies (i.e., makes a true statement of) the DE.
    Read the problem a little more closely. There is no restriction on y. What they did was break things up into two cases:
    Case 1: y ≠ 0, from which they obtained the exponentional function.
    Case 2: y = 0.
  13. Mar 14, 2013 #12
    Opps. My mistake.

    Here is what I should've said:

    Why is it implied that just because y=0 satisfies the DE, that y=0 also satisfies ##y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }##? In order to obtain ##y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }##, we had to restrict y=0 (basically case 1 that you referred to).

  14. Mar 14, 2013 #13


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    Science Advisor

    A is an arbitrary constant- taking any specific value for A gives a solution.

    So what if A= 0?
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