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Math Help - What sort of DE is this?

  1. #1
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    What sort of DE is this?

    \dot{x}=x-x^2-\lambda where \lambda \ge 0 is a constant.

    It's not Bernoulli, since it's not of the form

    \dot{x}= ax + bx^n

    So what is it? And how do I solve it?
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  2. #2
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    Oh, perhaps it's separable?

    So then I get \int \frac{1}{x-x^2-\lambda}~dx=\int 1 dt

    But how do I integrate the LHS? Partial fraction expansion? Anything easier?
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  3. #3
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    The Differential equation is a special equation called the Riccati equation. It reduces to the Bernoulli's equation for \lambda = 0

    The solution of this equation is not very simple. One method is to make the substitution x = u(t) + y

    where u(t) is a particular solution that is already known,
    y is a new dependent variable.

    This substitution reduces the equation to a Bernoulli's equation in y and t, and you can obtain y as a function of t. Returning to the original dependent variable gives you the actual solution as x(t) = u(t) + y

    In this case, you can take the particular solution to be one of the roots \alpha of the equation x^{2}-x+\lambda = 0 wherever \alpha is real.
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    Oh, perhaps it's separable?

    So then I get \int \frac{1}{x-x^2-\lambda}~dx=\int 1 dt

    But how do I integrate the LHS? Partial fraction expansion? Anything easier?
    The solution will depend on \lambda since x^2 - x + \lambda might have:

    Case 1: two distinct linear factors,

    Case 2: one repeated linear factor, or

    Case 3. no real linear factor

    depending on the value of \lambda. For case 1, use partial fractions. For case 2, it's a standard form. For case 3 you need to complete the square and then recognise a standard form that will give you arctan.

    The work hack work is left for you. (You might find it easier to start with concrete values for \lambda for each case. Note that there is only one value for case 2).
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  5. #5
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    Ah ok, so I think I have standard form, since I have lambda = 1/4 (I forgot to mention that part).

    So I have

    \int -1/(x-1/2)^2 ~dx = \int 1 ~dt

    Hmm.. How do I evaluate that integral?
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    Ah ok, so I think I have standard form, since I have lambda = 1/4 (I forgot to mention that part).

    So I have

    \int -1/(x-1/2)^2 ~dx = \int 1 ~dt

    Hmm.. How do I evaluate that integral?
    You're expected to know that \displaystyle \int (ax + b)^n \, dx = \frac{1}{a(n+1)} (ax + b)^{n+1} + C for n \neq -1.
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  7. #7
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    Oh, of course!
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