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Math Help - Please help ODE ( separable) one problem'''

  1. #1
    cyy
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    Please help ODE ( separable) one problem'''

    hi given (2x - 4y + 5 ) * ( dy/dx ) + x - 2y + 3 = 0


    I'v let u = x - 2y; dy/dx = 1-2*(dy/dx); dy/dx = 1/2 * ( 1 - du/dx )

    sub into origin eqn:
    ( 2u + 5 ) [ 1/2 ( 1-du/dx ) ] + u + 3 = 0

    ( 2u + 5 ) [ 1/2 - 1/2 * du/dx ] + u + 3 = 0
    ( 2u + 5 ) / 2 - ( ( 2u + 5 ) / 2 ) * du/dx = -u - 3
    ( 2u + 5 ) du/dx = 4u + 11 ....

    so how i should carry on from here? thx u!!!
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by cyy View Post
    hi given (2x - 4y + 5 ) * ( dy/dx ) + x - 2y + 3 = 0


    I'v let u = x - 2y; dy/dx = 1-2*(dy/dx); dy/dx = 1/2 * ( 1 - du/dx )

    sub into origin eqn:
    ( 2u + 5 ) [ 1/2 ( 1-du/dx ) ] + u + 3 = 0

    ( 2u + 5 ) [ 1/2 - 1/2 * du/dx ] + u + 3 = 0
    ( 2u + 5 ) / 2 - ( ( 2u + 5 ) / 2 ) * du/dx = -u - 3
    ( 2u + 5 ) du/dx = 4u + 11 ....

    so how i should carry on from here? thx u!!!
    So integrate now

    \int \frac{2u + 5}{4u + 11} \, du = \int dx

    \frac12\int \frac{4u + 11 - 1}{4u + 11} \, du = \int dx

    \frac12\int \left(1 - \frac{1}{4u + 11}\right) \, du = \int dx

    \frac{u}2 - \frac18 \ln|4u+11|  = x + C

    Resubstituting back:

    \frac{x - 2y}2 - \frac18 \ln|4x - 8y+11|  = x + C
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