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Math Help - Differential equation

  1. #1
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    Differential equation

    Hi I am trying to solve the following differential equation but not having much luck!

    dy/dx = 1+y^2 / 1 + x^2

    Its a multiple choice question and the answer is one of the following:

    (A) Cx / (1 - Cx)

    (B) Cx / (1 + Cx)

    (c) (C - x) / (1 - Cx)

    (D) (1 - Cx) / (x + C)

    (E) (x + C) / (1 - Cx)

    Here is what I have so far:

    Rearange to the form:

    1/(1+y^2) dy/dx = 1/ (1+x^2)

    Integrate both sides with respect to x, giving:

    tan^-1(y) = tan^-1(x) + c

    which is the same as:

    y = tan(tan^-1(x) + c)

    y = x + tan(c)

    we can let C = tan(c) as it is constant, giving:

    y = x + C

    but this looks nothing like the answer.

    Any help much appreciated, have spent ages trying to solve this!
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by alexthedon View Post
    y = tan(tan^-1(x) + c)

    y = x + tan(c)
    Here you messed up.
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  3. #3
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    I thought that...

    I thought that tan(tan^-1(x)) = x ?

    am I wrong there is is it something to do with the c?
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  4. #4
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    Yeah, but just works for that case.

    What about if you apply the formula \tan(\alpha+\beta)?
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  5. #5
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    Ahhh I see

    Many thanks.




    y = \tan(tan^-1(x) + c)

    Using:

    <br />
\tan(\alpha+\beta) <br />

    we get:

    tan(tan^-1(x)) + tan(c) / 1 - tan(tan^-1(x)) * tan(c)

    which gives us:

    x + tan(c) / 1 + x.tan(c)

    if we now let C = tan(c)

    we have:

    x + c / 1 - Cx


    I obviously need to know my Trig Cheat sheet better!

    Cheers,

    Alex
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