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Math Help - Find a Particular Solution, given initial condition

  1. #1
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    Find a Particular Solution, given initial condition

    (x+y)dx+(3x+3y-4)dy=o ---> y(1)=1

    x+y=u
    dy=du-dx

    3u-4=3x+3y-4

    udx+ (3u-u)(du-dx)=0

    Solving, we arrive at

    dx + (3u-4)/(4-2u)=0

    x- 1/2(2 ln(u-2)+ 3u)=c

    2x-(2 ln (u-2) + 3u)=c

    Substituting

    2x-2ln(x+y-2) - 3 (x+y)=c
    -x-3y-2ln(x+y-2)=c

    x+3y+2ln(x+y-2)=c

    So here are the problems I ran into:

    The answer in the book has the following answer:
    x+3y+2ln(2-x-y)=c


    If you plug in the Initial Value for x and y, you get

    1+3+ 2Ln(2-1-1)= 4+2Ln(0), which obviously creates a problem since LN(0) if undefined.

    Help Please!!!


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  2. #2
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    Let's continue a bit with your solution. So

    \ln 2 - x - y = \dfrac{c - x - 3y}{2}
    and exponentiating gives

    2 - x- y = e^{\frac{c-x-3y}{2}} = k e^{\frac{-x-3y}{2}}.

    Your IC of y(1) = 1 gives k = 0 and your solution is y = 2 - x.
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  3. #3
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    The answer in the book says that the general solution is

    x+3y+2 ln(2-x-y)=c.. the LN is taken of the whole expression in the parentheses. Where did the x and 3y go in your solution?

    So we're solving for c, not for y.

    The end result should be x+3y+2ln(2-x-y)=equals some value.

    Thanks for your help!
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  4. #4
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    The two answers

    x+3y+2 ln(2-x-y)=c

    2 - x - y = k e^{\dfrac{- x - 3y}{2}}

    are equivalent however yours cannot recover the initial condition where as mine can.

    This problem is similar to finding the solutions to

    \dfrac{dy}{dx} = y(1-y) subject to (i) y(0) = 0, and (ii) y(0) = 1.
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