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Solutions for x<0

  1. Dec 10, 2011 #1
    Hi,

    I have solved a differential equation (1) for x, where i think is a real variable.

    (1) [tex] \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1+x)^2} [/tex]

    And get a solution

    [tex] y(x) = \frac{2}{\sqrt{x}} - \frac{\sqrt{x}}{(1+x)} - 3 \arctan{\sqrt{x}} [/tex]

    Now, i want to find solutions for x<0, and i have said that the above is a solution for x>0.

    So far i have simply done a coord transformation x→-x and then solved this equation

    [tex] \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1-x)^2} [/tex]

    and then substituted in "-x" under my square roots.

    Is this a correct way of finding x<0?

    Konig.
     
    Last edited: Dec 10, 2011
  2. jcsd
  3. Dec 10, 2011 #2

    SammyS

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    What do you plan to do about the square root of negative numbers?
     
  4. Dec 10, 2011 #3
    So [tex] \sqrt{-x}, x<0 [/tex] will give a positive answer.

    Although im not sure what to do about the [tex] x^{\frac{3}{2}} [/tex] part of the equation.

    The answer should have natural logs and some x terms.

    the solution of [tex] \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1-x)^2} [/tex]

    gives a similar answer to what i need which gives me hope. But im not sure what the do with the first terms, eg there signs etc.

    thanks.
     
  5. Dec 10, 2011 #4

    SammyS

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    You can't arbitrarily change the sign of x under the radical.

    This will show the problem:
    [itex]\displaystyle\sqrt{x}=\sqrt{(-1)(-1)x}[/itex]
    [itex]\displaystyle =\left(\sqrt{-1}\,\right)\sqrt{-x}[/itex]​

    What will you do with [itex]\sqrt{-1}\,?[/itex] -- even if you call it i .
     
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