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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

    \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0

    Thanks
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  2. #2
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    Quote Originally Posted by Danog View Post
    Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

    \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0

    Thanks
    What is your aim? To find the second derivative of y?
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  3. #3
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    Agh, my bad. The questions asks to find the general solution to the equation, and to express y as a function of x.
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  4. #4
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    Quote Originally Posted by Danog View Post
    Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

    \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0

    Thanks
    \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0
    \frac{dy}{dx} = (3 + 2x)(1 + y^2)
    \frac{dy}{(1 + y^2)} =(3 + 2x) \cdot dx
    just integrate both side....
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  5. #5
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    Quote Originally Posted by Danog View Post
    Agh, my bad. The questions asks to find the general solution to the equation, and to express y as a function of x.
    This isn't really 'implicit differentiation', then. It is the solution of Ordinary Differential Equations!
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  6. #6
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    Quote Originally Posted by ramiee2010 View Post
    \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0
    \frac{dy}{dx} = (3 + 2x)(1 + y^2)
    \frac{dy}{(1 + y^2)} =(3 + 2x) \cdot dx
    just integrate both side....
    You use the lazy notation :P.

    To be COMPLETELY correct...


    \frac{dy}{dx} = (3 + 2x)(1 + y^2)

    \frac{1}{1 + y^2}\,\frac{dy}{dx} = 3 + 2x

    \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{3 + 2x\,dx}

    \int{\frac{1}{1 + y^2}\,dy} = 3x + x^2 + C_1

    \arctan{y} + C_2 = 3x + x^2 + C_1

    \arctan{y} = 3x + x^2 + C, where C = C_1 - C_2

    y = \tan{(3x + x^2 + C)}.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    You use the lazy notation :P.

    To be COMPLETELY correct...

    it was just a hint not a complete solution .
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  8. #8
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    Quote Originally Posted by ramiee2010 View Post
    it was just a hint not a complete solution .
    I'm aware of that. And it was correct. I'm just saying you were using lazy shortcuts that many mathematicians would vomit on...
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