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Math Help - Integration by Separating variables

  1. #1
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    Integration by Separating variables

    ok so i have the differential equation:

    dx/dt = ((x-1)^2)*(t+1)

    this is separable and separation of variables gives

    integral ( dx/((x-1)^2)) = integral (t+1) dt

    im told the correct answer is..

    -1 / (x-1) = ((t+1)^2)/2 + c

    but am wondering, why cant i go back to the differential equation and integrate the RHS as:

    (t^2)/2 + t + c

    why is (t+1) integrated together... why not integrate t first... and then 1... ? surely thats more simple??
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  2. #2
    Senior Member Shanks's Avatar
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    They make no difference, you will get the same result.
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  3. #3
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    maybe im the only one not seeing it,when i integrate the t and 1 separately i dont get an extra 1/2 .... which i get when i integrate (t+1) together...

    im sure this is v.basic but im not seeing it!
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  4. #4
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    Quote Originally Posted by matlabnoob View Post
    ok so i have the differential equation:

    dx/dt = ((x-1)^2)*(t+1)

    this is separable and separation of variables gives

    integral ( dx/((x-1)^2)) = integral (t+1) dt

    im told the correct answer is..

    -1 / (x-1) = ((t+1)^2)/2 + c

    but am wondering, why cant i go back to the differential equation and integrate the RHS as:

    (t^2)/2 + t + c

    why is (t+1) integrated together... why not integrate t first... and then 1... ? surely thats more simple??

    \int (t+1) dt = \frac{t^2}{2} + t + C ( I would also say that this way is more simple)

    but also

    \frac{(t+1)^2}{2} = \frac {t^2 + 2t +1}{2} + C = \frac{t^2}{2} + t + \frac{1}{2} + C = \frac{t^2}{2} + t + C

    1/2 is a constant
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  5. #5
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    thanks!!
    1/2 is a constant...of course..!

    believe it or not i am doing a maths degree how embarrassing
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