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Math Help - dy/dx=x^2y-y^2+2x

  1. #1
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    dy/dx=x^2y-y^2+2x

    Here's the problem:

    the solution to y0 is dy/dx=x^2y-y^2+2x; and y(-2)=3-- now we are also told that y'0(-2)=1 now for the part im having trouble with: find y''0(-2)
    now from the problem i was thinking to take the derivative of the equation they gave me... so that would mean implicit differentiation??: 2xy+x^2y'-2yy'+2... but when i plug in -2 for x and 3?? for y i get -5... any help is appreciated ty (Nod)
    Last edited by mr fantastic; March 3rd 2011 at 12:32 PM. Reason: Moved and re-titled.
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  2. #2
    MHF Contributor
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    Hi

    y'' = 2xy + x^2y' - 2yy' + 2

    y_0''(-2) = 2(-2)y_0(-2) + (-2)^2y_0'(-2) - 2y_0(-2)y_0'(-2) + 2

    You need to plug in -2 for x, 3 for y0(-2) and 1 for y0'(-2)
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  3. #3
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    Hey, thanks for reply. Ok it hadn't occured to me about plugging in 1 for y'. I also realized I made a typo since it was supposed to be y0'(-2)=-1 which comes out to be -8 instead of -12 of what I was getting when I plugged in 1. Thanks again lol
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