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Intro to diffrential equn simple question

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data

    hey, it's been on my wish list for some time, i have decided to teach my self diffrential equal, rather than waiting to be taught at school, after having some exposure through vibrations and circuits.

    so i got a book and i started to learn, the question came to be;

    [tex]x^{3}[/tex] [tex]\frac{dy}{dx}[/tex] = y





    2. The attempt at a solution
    so i solve it by seperation of variable and arrived at the answer of

    y=[tex]e^{-.5x^{2}+c}[/tex]

    i am afride it is wrong, or am i just confused.
     
  2. jcsd
  3. May 4, 2010 #2

    lanedance

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    Homework Helper

    try and show your working, and note you can put a whole equation in tex tags
    [tex]x^3 \frac{dy}{dx} = y[/tex]

    did you separate like below?
    [tex] \frac{dy}{y} = \frac{dx}{x^3} [/tex]
     
  4. May 4, 2010 #3
    yes.
     
  5. May 4, 2010 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Then you've lost a sign.

    If [itex]dy/y= dx/x^3= x^{-3}dx[/itex] then

    [tex]ln(y)= -(1/2)x^{-2}+ C[/tex]
    and so

    [tex]y= e^{-.5x^{-2}+ C[/tex]

    It should be [itex]x^{-2}[/itex] in the exponent, not [itex]x^2[/itex].
     
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