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Math Help - Antiderivatives

  1. #1
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    Antiderivatives

    How do you find the antiderivatives of:

    1. x+4y^(3)*[(x^(2)+1)^2]*dy/dx=0

    2. (y^2)/(x^6)=dy/dx

    3. y^2+1=dy/dx

    4. y'=(sinx)/(siny)

    Thanks!
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  2. #2
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    differential equations

    Well, they are some differential equations.
    In fact,they are all separable differential equations.
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    Last edited by curvature; June 5th 2007 at 11:42 PM.
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  3. #3
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    Quote Originally Posted by sw3etazngyrl View Post
    How do you find the antiderivatives of:

    1. x+4y^(3)*[(x^(2)+1)^2]*dy/dx=0

    ...
    Hello,

    x+4y^3 \cdot (x^2+1)^2 \cdot \frac{dy}{dx}=0\; \Longleftrightarrow \; 4y^3 \cdot dy = -\frac{x}{(x^2+1)^2} \cdot dx

    \int 4y^3 \cdot dy = \int \left(-\frac{x}{(x^2+1)^2} \right) \cdot dx For the RHS use integration by substitution:

    y^4 = \frac{1}{2(x^2+1)} + C \; \Longrightarrow \; y = \left|\sqrt[4]{\frac{1}{2(x^2+1)} + C}\; \; \right|
    Last edited by earboth; June 6th 2007 at 04:33 AM.
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  4. #4
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    Quote Originally Posted by earboth View Post
    Hello,

    x+4y^3 \cdot (x^2+1)^2 \cdot \frac{dy}{dx}=0\; \Longleftrightarrow \; 4y^3 \cdot dy = -\frac{x}{(x^2+1)^2} \cdot dx

    \int 4y^3 \cdot dy = \int \left(-\frac{x}{(x^2+1)^2} \right) \cdot dx For the RHS use integration by substitution:

    y^4 = \frac{1}{2(x^2+1)} + C \; \Longrightarrow \; y = \left|\sqrt[4]{\frac{1}{2(x^2+1)} + C}\; \; \right|
    hi
    was wondering if there are any constant?
    edit
    ok its fixed
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  5. #5
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    No 1

    the solution of the 1st question again
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