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Math Help - Help for solving the differential equation!!

  1. #1
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    Help for solving the differential equation!!

    The Qs is to solve the differential equation:
    \frac{dy}{dx} = \frac{(x+y)}{(x-y)} with y(1) = 1
    I have tried to use Homogeneous differential equation rules:
    \frac{dy}{dx} = H(\frac{y}{x})
    Then y(x) = xu(x); x\frac{du}{dx} + y = H(u)

    Relative to the real Qs, my step is shown below
    \frac{dy}{dx} = \frac{x+y}{x-y} = .... = \frac{1}{1-\frac{y}{x}} + \frac{1}{[\frac{1}{(\frac{y}{x})}-1]} = H(\frac{y}{x})
    By Homogeneous differential equation rules
    I got \frac{du}{dx} = \frac{1+u^2}{(1-u)x}
    and after I antidifferentiate this equation
    arctan(u) - 0.5log(1+u^2) = log|x|+C
    after that, I don't know how to continue, please help!
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  2. #2
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    Krizalid's Avatar
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    you're making it too complicated.

    note the RHS, divide top and bottom by x and then put y=ux.
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  3. #3
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    even I use your method the answer for du/dx won't change, and so I'll get the same solution as before, the most important thing is the step after I found du/dx.
    Please provide more detail for the next step!!
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  4. #4
    Bar0n janvdl's Avatar
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    \frac{dy}{dx} = \frac{x+y}{x-y} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}}

    Let u = \frac{y}{x} then dy = x \ du + u \ dx

    Now substitute the above back into the original equation.

    x \frac{du}{dx} + u = \frac{1 + u}{1 - u}

    x \frac{du}{dx} = \frac{1 + u^2}{1 - u}

    Thus:

    \int \frac{1 - u}{1 + u^2} du = \int \frac{dx}{x}

    \int \frac{1}{1 + u^2} - \int \frac{u}{1 + u^2} = \ln{(x)} + C

    \arctan(u) - \frac{1}{2} \ln{(u)} = \ln{(x)} + C

    But u = \frac{y}{x}

    \arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln{\left( \frac{y}{x} \right)} = \ln{(x)} + C
    Last edited by janvdl; September 24th 2009 at 09:00 AM. Reason: Found a mistake, and corrected it.
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  5. #5
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    You could solve them simultaneously:

    \frac{dx}{dt}=x+y

    \frac{dy}{dt}=x-y

    calculte eigenvalues and come up with parametric solutions which because of the complex eigenvalues are:

    x(t)=c_1 e^{t}\cos(t)+c_2e^t\sin(t)

    y(t)=c_3 e^{t}\cos(t)+c_4e^t\sin(t)

    but the arbitrary constants are related so only have two for an IVP. It's a spirial.
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  6. #6
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    Quote Originally Posted by janvdl View Post
    \frac{dy}{dx} = \frac{x+y}{x-y} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}}

    Let u = \frac{y}{x} then dy = x \ du + u \ dx

    Now substitute the above back into the original equation.

    x \frac{du}{dx} + u = \frac{1 + u}{1 - u}

    x \frac{du}{dx} = \frac{1 + u^2}{1 - u}

    Thus:

    \int \frac{1 - u}{1 + u^2} du = \int \frac{dx}{x}

    \int \frac{1}{1 + u^2} - \int \frac{u}{1 + u^2} = \ln{(x)} + C

    \arctan(u) - \frac{1}{2} \ln{(u)} = \ln{(x)} + C

    But u = \frac{y}{x}

    \arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln{\left( \frac{y}{x} \right)} = \ln{(x)} + C
    I think \arctan(u) - \frac{1}{2} \ln{(1+u^2)} = \ln{(x)} + C should be right
    But not \arctan(u) - \frac{1}{2} \ln{(u)} = \ln{(x)} + C
    And so we'll got \arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln{\left(1+( \frac{y}{x})^2 \right)} = \ln{(x)} + C finally.

    Then, the thing make me confused is how to transfer this equation to y=f(x), that looks like hard. Please help!
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by wsun View Post
    I think \arctan(u) - \frac{1}{2} \ln{(1+u^2)} = \ln{(x)} + C should be right
    But not \arctan(u) - \frac{1}{2} \ln{(u)} = \ln{(x)} + C
    And so we'll got \arctan \left( \frac{y}{x} \right) - \frac{1}{2} \ln{\left(1+( \frac{y}{x})^2 \right)} = \ln{(x)} + C finally.

    Then, the thing make me confused is how to transfer this equation to y=f(x), that looks like hard. Please help!
    Yes you are right. Just swap that (u) for (1 + u^2).

    Normally your lecturer would not expect you to put it in the form y = f(x).
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