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Math Help - 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  y^2 \ dy = 3x^{5} \ dx .

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


    Or  2y^3 = 3x^6 + 6c

    Let  6c = C and we get

     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:

    <br />
y^2dy - 3x^5dx = 0<br />

    so:

    <br />
\int y^2dy - \int 3x^5dx = K<br />

    <br />
\frac{y^3}{3} - \frac{x^6}{2} = K<br />

    Now multiply through by 6:

    <br />
2y^3 - 3x^6 = C<br />

    (the constant K has now been replaced by the constant C=6K)

    Reaarange and you have the result.

    RonL
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