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Math Help - Differential Equation y' + y cotx = sinx

  1. #1
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    Differential Equation y' + y cotx = sinx

    Hi guys, this is the differentioan equation I need help to solve

    y' + y cotx = sinx

    attempt:
    '
    This is the case of using an Integration factor as far as I am concerned.

    The form of the equation is

    dy/dx (with constant 1) + a(x) * y = b(x)


    1. I can see that a(x) is in fact cot x . my integration factor will be e^$cot x dx

    as cot x is the same as 1/tanx I put it in the I.F instead

    Thus e^$1/tanx * dx -> e^ln tan x = I.F = tanx

    2. Then i multiply it

    tanx * y' + tanx * y cot x = tanx * sinx


    as far as I can see, this is not the product rule?

    Can someone help me solve this?
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  2. #2
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    $ 1/x = ln x (true)

    but for $ 1/tanx you would need the chain rule - in fact, think of it as $ cos x / sin x...

    Just in case a picture helps...

    Spoiler:



    ... do you have the IF?

    Key:
    Spoiler:



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    If the picture helps, use it for the IF, too, via something like...



    See http://www.ballooncalculus.org/examp...ce.html#factor

    Last edited by tom@ballooncalculus; February 26th 2011 at 02:52 PM.
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  3. #3
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    Isn't the primitive function of cot x = log sinx, or is it the same as ln?
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  4. #4
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    That's right... ln = log base e = natural log
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  5. #5
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    so sinx is the I.F --->

    sinx * y' + sinx * y* cosx/sinx
    = sinx^2

    sinx * y' + y*cosx = sinx^2

    (sinx * y)' = (sinx)^2

    After integrating

    (sinx * y) = 1/2 * (x-sinx*cosx)

    y = 1/2 * (x-sinx*cosx)
    --------------------
    sinx


    [B]Wrong or right? could you possibly solve it for me?[B]
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  6. #6
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    All good but of course don't forget to + c
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