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Solution to differential equation

  1. May 24, 2013 #1
    Obtain the solution to the differential equation:

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

    Multiple choice answer:

    a) [tex]\frac{Cx}{1-Cx}[/tex]
    b) [tex]\frac{Cx}{1+Cx}[/tex]
    c) [tex]\frac{C-x}{1-Cx}[/tex]
    d) [tex]\frac{1-Cx}{x+C}[/tex]
    e) [tex]\frac{x+C}{1-Cx}[/tex]

    Tried integrating two sides to arrive at arctan y = arctan x + C, but not sure how to proceed from there.
     
  2. jcsd
  3. May 24, 2013 #2

    CAF123

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    Rearrange first to get $$\tan^{-1}y - \tan^{-1}x = C$$ Then take tan on both sides and simplify.
     
  4. May 24, 2013 #3
    I'm going to write the simplification out, because it took me a while.

    [tex]\tan^{-1}y - \tan^{-1}x = C[/tex]

    In order to take tan on the left side, this equation needs to be re-written as:

    [tex]\tan^{-1}\frac{y-x}{1+yx}=C[/tex]

    then we can take the tangent of both sides, giving us:

    [tex]\frac{y-x}{1+yx}=C[/tex]

    then we solve for y:

    [tex]{y-x}={C+Cyx}[/tex]

    [tex]0 = C+Cyx-y+x[/tex]

    [tex]0 = y(Cx-1)+C+x[/tex]

    [tex]y = \frac{-C-x}{Cx-1}[/tex]

    [tex]y = \frac{x+C}{1-Cx}[/tex]

    Answer E above
     
    Last edited: May 24, 2013
  5. May 24, 2013 #4

    CAF123

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    Gold Member

    I agree with the answer E, but you can take tan from the beginning: $$ \tan (\tan^{-1}y - \tan^{-1}x) = \tan K$$ Now let ##\tan^{-1}y = \theta,\,\,\,\tan^{-1}x = \phi##, so we have $$\tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi},$$ using addition formulae. Sub in the above conditions and let ##\tan K = C## then rearrange gives the result.
     
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