What sort of DE is this?

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October 31st 2010, 07:48 PM
scorpion007

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?
October 31st 2010, 08:49 PM
scorpion007

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?
October 31st 2010, 10:36 PM
bandedkrait

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.
October 31st 2010, 10:42 PM
mr fantastic

Quote:

Originally Posted by scorpion007

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).
November 1st 2010, 12:29 AM
scorpion007

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?
November 1st 2010, 01:39 AM
mr fantastic

Quote:

Originally Posted by scorpion007

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 $n \neq -1$ .
November 1st 2010, 02:24 AM
scorpion007

Oh, of course!