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A first order ODE

  1. Sep 22, 2005 #1
    I need to solve the following for f(s):

    (s^2 + 1)f '(s) + s f(s) = 0

    First I isolated for f '(s), which gave me:

    f '(s) = -s f(s)/(s^2 + 1)

    Then,

    d f(s)/ds = -s f(s)/(s^2 + 1)

    so, d f(s) = (-s f(s)/(s^2 + 1))ds

    Integrating I get:

    f(s) = -F(s)ln(s^2 + 1)/2 + C, where F(s) is the antiderivative of f(s) and C is a constant of integration.

    Did I do any of that correctly??!!??
     
  2. jcsd
  3. Sep 22, 2005 #2

    saltydog

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    Not the last part. Just write it this way:

    [tex]f^{'}+\frac{s}{s^2+1}f=0[/tex]

    That's in standard form right? Now calculate the integrating factor [itex]\sigma [/itex], multiply both sides by it, end up with an exact differential on the LHS, integrate, don't forget the constant of integration, that should do it. This is the integrating factor:

    [tex]\sigma=\text{Exp}\left[\int \frac{s}{(s^2+1)}\right][/tex]

    Can you do the rest?
     
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