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Why is this not a solution to the differential equation

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Explain why the piecewise-defined function:
    [itex]y=\begin{Bmatrix}
    \sqrt{25-x^{2}}, & -5< x< 0 \\
    -\sqrt{25-x^{2}},&0\leq x< 5
    \end{Bmatrix}[/itex]

    is not a solution of the differential equation [itex]\frac{dy}{dx}=-\frac{x}{y}[/itex] on the interval (-5,5)


    2. The attempt at a solution
    I have no idea why...Differentiating y i get
    [itex]y'=\begin{Bmatrix}
    \frac{-x}{\sqrt{25-x^{2}}}, & -5< x< 0 \\
    \frac{x}{\sqrt{25-x^{2}}},&0\leq x< 5
    \end{Bmatrix}[/itex]

    Substituting the denominators in y' by the corresponding expression defined by y, I get back exactly [itex]y'=-\frac{x}{y}[/itex] for the two pieces of interval.

    To me it should be a solution...

    Thank you very much.

    EDIT:
    The question comes from here (#26):
    http://books.google.ca/books?id=qh1...ifferential equation on the interval"&f=false
     
    Last edited: Sep 10, 2011
  2. jcsd
  3. Sep 10, 2011 #2
    Well, a solution has to be differentiable at the segment... This function isn't even continuous...
     
  4. Sep 10, 2011 #3
    Hmm I don't really understand... I went back to read the definition of a solution and it said:

    So, from what I understand, if F is a solution then its n derivatives must be continuous on I.

    [itex]y'=\begin{Bmatrix}
    \frac{-x}{\sqrt{25-x^{2}}}, & -5< x< 0 \\
    \frac{x}{\sqrt{25-x^{2}}},&0\leq x< 5
    \end{Bmatrix}[/itex]

    is continuous on I since lim as x goes to 0 of y' is 0 in both pieces.
     
    Last edited: Sep 10, 2011
  5. Sep 10, 2011 #4

    vela

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    What's the derivative at x=0? Hint: It's not 0.
     
  6. Sep 10, 2011 #5
    Why not? I would have said 0... x/sqrt(25-x^2) = 0/sqrt(25) = 0...
     
  7. Sep 10, 2011 #6

    vela

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    Try plotting the function. You'll see why.
     
  8. Sep 10, 2011 #7
    Here is the plot: http://img856.imageshack.us/img856/5156/function2h.th.jpg [Broken]

    Intuitively I would have said that the derivative at x=0 is 0 since as x goes to 0 both pieces have a slope of 0 by looking at the graph.

    However, now I remember that by definition a function must be continuous to be differentiable. I don't really understand why though.

    Consider y=x^2 which we break in two at x=0 and move one part 5 units up:http://img40.imageshack.us/img40/2070/function1.th.jpg [Broken]

    Even though it has a discontinuity, it don't see how the derivative is affected by this modification... it is still 2x on all of the interval, even 0....
     
    Last edited by a moderator: May 5, 2017
  9. Sep 10, 2011 #8

    vela

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    For both functions, the slope approaches 0 near the origin, but it's undefined at the origin. Go back to the definition of the derivative:
    [tex]f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}[/tex]
    For that limit to exist, the left-handed and right-handed limits have to agree, but one of them diverges.
     
  10. Sep 10, 2011 #9

    LCKurtz

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    Here's why. If f is differentiable at x0 then
    [tex]\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}=f'(x_0)[/tex]

    Intuitively, since h → 0 the only way the fraction can have a limit is if the numerator → 0, which says f is continuous at x0. To prove it rigorously write
    [tex]\frac{f(x_0+h)-f(x_0)}{h}=f'(x_0)+\left(\frac{f(x_0+h)-f(x_0)}{h}-f'(x_0)\right)
    =f'(x_0) + o(h)[/tex]
    where o(h) represents the quantity in parentheses which → 0 as h → 0 since f is differentialbe at x0. Then
    [tex]f(x_0+h)-f(x_0) = hf'(x_0) + ho(h)[/tex]

    Now if you let h → 0 it tells you that f(x0+h)→ f(x0) as h → 0 so f is continuous at x0.
     
    Last edited: Sep 10, 2011
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