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Math Help - help with a problem

  1. #1
    Junior Member
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    help with a problem

    Hey guys, i just started a new course in differential equations and im kinda having a hard time. Heres one of the problems I am having difficulty with:

    Verify that the piecewise-defined function

     y=\left\{\begin{array}{cc}-x^2,&\mbox{ if }<br />
x < 0\\x^2, & \mbox{ if } x\geq 0\end{array}\right.

    is a solution of the differential equations  xy'-2y = 0 on  (-\infty , \infty)

    Thanks!
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  2. #2
    A Plied Mathematician
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    What have you tried so far?
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  3. #3
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    Im not even sure how to start this problem. Is there a clue you can give me?
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  4. #4
    A Plied Mathematician
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    Well, you're not asked to solve the DE, but to verify the solution. So, I would plug the given solution into the DE and see if you get an equality. What do you get when you follow that path?
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  5. #5
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    hmph. i just did:
    if  x = -1

     -2x^2 +2x^2 = 0

     -2 + 2 = 0

     0 = 0

     if x = 1

     2x^2 -2x^2 = 0

     2 - 2 = 0

     0=0

    Is this what its asking for?
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  6. #6
    MHF Contributor

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    Not quite. You cannot just choose two values of x. If x is any positive number, then y= x^2 so that y'= 2x. The equation xy'-2y = 0 becomes x(2x)- 2x^2= 2x^2- 2x^2= 0 so the equation is satisfied for any positive x. If x is negative, y= -x^2. Do the same with that. Most importantly, what happens at x= 0? In particular what is the derivative of y at x= 0? (You can't just differentiate x^2 and set x= 0 because the derivative involves a limit from both sides: If x= 0 and h is positive, then x+h= 0+ h= h is positive, if h is negative, x+ h= h is negative.

    \lim_{h\to 0^+} \frac{y(h)- y(0)}{h}= \lim_{h\to 0}\frac{2h^2}{h}= ?
    \lim_{h\to 0^-} \frac{y(h)- y(0)}{h}= \lim_{h\to 0}\frac{-2h^2}{h}= ?
    If those two limits are the same, that is the dervative at x= 0. If they are not, the function is not differentiable at x= 0.
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