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    • Thread Starter
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    du/dt = au(b-u).

    By separating variables and using partial fractions show that

    ln (u/(b-u)) = abt + C

    If the solution didnt have the b on the RHS i would be okay, but I cannot see how to do this with the b on the RHS. Any help appreciated.

    Zezzy
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    Use partial fractions
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    (Original post by Serano)
    Use partial fractions
    How on this question?
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    show your working so far...
    and yeh doing some re-arranging in my head, i'd probably use partial fractions too
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    (Original post by didgeridoo12uk)
    show your working so far...
    du/dt = au(b-u)

    du/ u(b-u) = a dt

    but I know I need something like 1/u + 1/(b-u) = ab dt to get the answer :/
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    (Original post by Zezzy)
    du/dt = au(b-u)

    du/ u(b-u) = a dt

    but I know I need something like 1/u + 1/(b-u) = ab dt to get the answer :/
    dont worry about trying to get the answer they've given you. just throw maths at it and worry about that when you've finished.

    so obviously not you need to do some partial fractions on 1/(u(b-u)) to split it up...
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    (Original post by didgeridoo12uk)
    dont worry about trying to get the answer they've given you. just throw maths at it and worry about that when you've finished.

    so obviously not you need to do some partial fractions on 1/(u(b-u)) to split it up...
    So like D/u + E/(b-u) = 1

    D(b-u) + E(u) = 1

    then what?
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    (Original post by Zezzy)
    So like D/u + E/(b-u) = 1

    D(b-u) + E(u) = 1

    then what?
    substitue values in for u that will let you work out the values of D and E by eliminating either D or E.

    for example u = b
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    I have it! Thanks very much
 
 
 
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