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Thread: What sort of DE is this?

  1. #1
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    What sort of DE is this?

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

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

    $\displaystyle \dot{x}= ax + bx^n$

    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$\displaystyle \int \frac{1}{x-x^2-\lambda}~dx=\int 1 dt$

    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 $\displaystyle \lambda = 0$

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

    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 $\displaystyle x(t) = u(t) + y$

    In this case, you can take the particular solution to be one of the roots $\displaystyle \alpha$ of the equation $\displaystyle x^{2}-x+\lambda = 0$ wherever $\displaystyle \alpha$ 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$\displaystyle \int \frac{1}{x-x^2-\lambda}~dx=\int 1 dt$

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

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

    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

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

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