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Math Help - help w/ difficult IVP

  1. #1
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    help w/ difficult IVP

    Use separation of variables to obtain the solution to the following IVP:


    I can't seem to finish this one.

    \int dy/(y^2-4) = t + C

    ln(y-2) - ln(y+2) = 4(t+C)

    (y-2)/(y+2)=e^(4t+4C))

    solving for y gave me:

    y= (2e^(4t+4C)))/(1-e^(4t+4C)))

    so yeah its not turning out well for me
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  2. #2
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    Quote Originally Posted by cdlegendary View Post
    Use separation of variables to obtain the solution to the following IVP:


    I can't seem to finish this one.

    \int dy/(y^2-4) = t + C

    ln(y-2) - ln(y+2) = 4(t+C)

    (y-2)/(y+2)=e^(4t+4C))

    solving for y gave me:

    y= (2e^(4t+4C)))/(1-e^(4t+4C)))

    so yeah its not turning out well for me
    Unless someone else can see an easy method then this is just the way I would have done it.

    Except you're missing a bit in you y = ... part.

    I think it should be...

    y = \frac{2(e^{4t+4C} + 1)}{1-e^{4t+4C}}

    Then for the initial condition you get...

    \frac{2(e^{4C} + 1)}{1-e^{4C}} = 0

    Solving this gives... C = \frac{1}{4}i \pi.

    Hence we get y = \frac{2(1 - e^{4t})}{1+e^{4t}}
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  3. #3
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    ah yeah I mistyped part of my equation, but this method sems to still give the wrong answer
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  4. #4
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    Quote Originally Posted by cdlegendary View Post
    Use separation of variables to obtain the solution to the following IVP:


    I can't seem to finish this one.

    \int dy/(y^2-4) = t + C

    ln(y-2) - ln(y+2) = 4(t+C)

    (y-2)/(y+2)=e^(4t+4C))

    solving for y gave me:

    y= (2e^(4t+4C)))/(1-e^(4t+4C)))

    so yeah its not turning out well for me
    \frac{dy}{dx} = y^2 - 4

    \frac{dx}{dy} = \frac{1}{y^2 - 4}

    \frac{dx}{dy} = \frac{1}{(y - 2)(y + 2)}.


    Now using partial fractions:

    \frac{A}{y - 2} + \frac{B}{y + 2} = \frac{1}{(y - 2)(y + 2)}

    \frac{A(y + 2) + B(y - 2)}{(y - 2)(y + 2)} = \frac{1}{(y - 2)(y + 2)}.

    So A(y + 2) + B(y - 2) = 1

    Ay + 2A + By - 2B = 1

    (A + B)y + 2A - 2B = 0y + 1.

    So A + B = 0 and 2A - 2B = 1.

    From the first equation, we can see that B = -A.

    Therefore 2A - 2(-A) = 1

    4A = 1

    A = \frac{1}{4}.

    So B = -\frac{1}{4}.


    Therefore:

    \frac{1}{(y - 2)(y + 2)} = \frac{1}{4(y - 2)} - \frac{1}{4(y + 2)}.


    So \frac{dx}{dy} = \frac{1}{4(y - 2)} - \frac{1}{4(y + 2)}

    x = \frac{1}{4}\int{\frac{1}{y - 2}\,dy} - \frac{1}{4}\int{\frac{1}{y + 2}\,dy}

    x = \frac{1}{4}\ln{|y - 2|} - \frac{1}{4}\ln{|y + 2|} + C

    x = \frac{1}{4}\ln{\left|\frac{y - 2}{y + 2}\right|} + C

    x - C = \frac{1}{4}\ln{\left|1 - \frac{4}{y + 2}\right|}

    4x - 4C = \ln{\left|1 - \frac{4}{y + 2}\right|}

    e^{4x - 4C} = \left|1 - \frac{4}{y + 2}\right|

    e^{-4C}e^{4x} = \left|1 - \frac{4}{y + 2}\right|

    Ae^{4x} = 1 - \frac{4}{y + 2} where A = \pm e^{-4C}

    \frac{4}{y + 2} = 1 - Ae^{4x}

    \frac{y + 2}{4} = \frac{1}{1 - Ae^{4x}}

    y + 2 = \frac{4}{1 - Ae^{4x}}

    y = \frac{4}{1 - Ae^{4x}} - 2.


    Since y(0) = 0

    0 = \frac{4}{1 - Ae^{4(0)}} - 2

    2 = \frac{4}{1 - A}

    \frac{1}{2} = \frac{1 - A}{4}

    2 = 1 - A

    A = -1.



    Therefore y = \frac{4}{1 + e^{4x}} - 2

    y = \frac{4}{1 + e^{4x}} - \frac{2(1 + e^{4x})}{1 + e^{4x}}

    y = \frac{4 - 2 - 2e^{4x}}{1 + e^{4x}}

    y = \frac{2 - 2e^{4x}}{1 + e^{4x}}.
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