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Thread: 9. What method I can use on this equation?

  1. #1
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    9. What method I can use on this equation?

    It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

    Question
    $\displaystyle a^2(dy-dx) = x^2 dy + y^2 dx $ ; $\displaystyle a $ is constant

    Answer
    $\displaystyle 2 $ $\displaystyle tan^{-1} $ $\displaystyle \left(\frac{y}{a}\right) $ $\displaystyle = ln \left | \frac {c(x+a)}{(x-a)} \right | $
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  2. #2
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    Quote Originally Posted by redfox2600 View Post
    It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

    Question
    $\displaystyle a^2(dy-dx) = x^2 dy + y^2 dx $ ; $\displaystyle a $ is constant

    Answer
    $\displaystyle 2 $ $\displaystyle tan^{-1} $ $\displaystyle \left(\frac{y}{a}\right) $ $\displaystyle = ln \left | \frac {c(x+a)}{(x-a)} \right | $
    It clearly re-arranges into $\displaystyle (a^2 - x^2) dy = (y^2 + a^2) dx$ and is therefore seperable.

    If you need more help, please show your work.
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  3. #3
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    Thank mr fantastic,

    Here is my partial solution.




    Separation of the variables:


    $\displaystyle \frac {dy}{(a^2 + y^2)} = \frac {dx}{(a^2 - x^2)} $

    $\displaystyle \frac {dy}{(a^2 + y^2)} = \frac {dx}{(a + x)(a - x)} $


    Right side of the equation:

    $\displaystyle \int \frac {dx}{(a + x)(a - x)} $

    $\displaystyle \int \frac {dx}{(a + x)} \cdot \int \frac {dx}{(a-x)} $

    $\displaystyle \int \frac {dx}{(x+a)} \cdot \int \frac {dx}{-(x-a)} $

    $\displaystyle \int \frac {dx}{(x+a)} \cdot -\int \frac {dx}{(x-a)} $

    $\displaystyle ln \mid x + a \mid $ $\displaystyle -ln \mid x - a \mid +ln \mid c \mid $

    $\displaystyle ln \mid x + a \mid $ $\displaystyle - ln \mid x - a \mid +ln \mid c \mid $

    $\displaystyle ln \mid x + a \mid $ $\displaystyle + ln \left( \mid x - a \mid \right)^{-1} +ln \mid c \mid $

    $\displaystyle ln \left| \frac {(x + a)}{(x - a)} \right| +ln \mid c \mid $

    $\displaystyle ln \left| \frac {c(x + a)}{(x - a)} \right| $

    Did I wrote the equation correctly? Correct me if I am wrong.


    Left side of the equation:

    $\displaystyle \int \frac {dy}{(a^2 + y^2)} $
    I'm stucked in this point. How can I go to the answer (left side of the equation)?

    Answer:
    $\displaystyle 2 tan^{-1} \frac {y}{a} = ln \left| \frac {c(x+a)}{(x-a)} \right| $


    I think this theorem would be help. But I am not sure how to go to the answer.
    Theorem: Inverse Trigonometric Functions
    $\displaystyle \frac {d}{dx} (tan^{-1}) = \frac {1}{1 + u^2} \frac {du}{dx} $


    Thanks to all.
    Regards,
    Last edited by redfox2600; Jul 26th 2009 at 05:42 PM.
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  4. #4
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    Quote Originally Posted by redfox2600 View Post
    $\displaystyle
    \int \frac {dx}{(x+a)} \cdot -\int \frac {dx}{(x-a)}
    $
    $\displaystyle ln \mid x + a \mid $ $\displaystyle -ln \mid x - a \mid +ln \mid c \mid $
    This is wrong. It's multiplication, not substraction
    Use partial fraction

    Left side of the equation:

    $\displaystyle \int \frac {dy}{(a^2 + y^2)} $
    I'm stucked in this point. How can I go to the answer (left side of the equation)?
    use substitution : y = a (tan theta)
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  5. #5
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    thanks songoku.



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