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Thread: differential equations

  1. #1
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    differential equations

    How would you do this:
    solve:
    y^2dy - 3x^5dx = 0
    the answer in the book is 2y^3 = 3x^6 + C
    I am clueless how to do these types of problems. The book doesnt explain it clearly and my notes arent helping me. Thanks.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    So $\displaystyle y^2 \ dy = 3x^{5} \ dx $.

    Then $\displaystyle \frac{y^3}{3} = \frac{x^6}{2} + c $


    Or $\displaystyle 2y^3 = 3x^6 + 6c $

    Let $\displaystyle 6c = C $ and we get

    $\displaystyle 2y^3 = 3x^6 + C $

    This is also separation of variables.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by davecs77 View Post
    How would you do this:
    solve:
    y^2dy - 3x^5dx = 0
    the answer in the book is 2y^3 = 3x^6 + C
    I am clueless how to do these types of problems. The book doesnt explain it clearly and my notes arent helping me. Thanks.
    Exactly like the other one:

    $\displaystyle
    y^2dy - 3x^5dx = 0
    $

    so:

    $\displaystyle
    \int y^2dy - \int 3x^5dx = K
    $

    $\displaystyle
    \frac{y^3}{3} - \frac{x^6}{2} = K
    $

    Now multiply through by 6:

    $\displaystyle
    2y^3 - 3x^6 = C
    $

    (the constant $\displaystyle K$ has now been replaced by the constant $\displaystyle C=6K$)

    Reaarange and you have the result.

    RonL
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