Nonlinear Autonomous System

**thomas_donald** · May 1st 2010, 01:14 PM

Can I get help on the solution of the following 2 systems:
1. dx/dt = -y(y-1) dy/dt = (x-1)(y-1)
2. dx/dt = 1/y dy/dt = 1/x
Thanka a lot for any help.

**kjchauhan** · May 1st 2010, 10:11 PM

(1) Pl Note that,

$\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx}$

$\therefore \frac{dy}{dx}=\frac{(x-1)(y-1)}{-y(y-1)}$

$\therefore y \cdot dy= -(x-1) \cdot dx$

$\therefore y \cdot dy+(x-1) \cdot dx = 0$

Now integrate..

Same way,,,

(2)

$\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx}$

$\therefore \frac{dy}{dx}=\frac{y}{x)}$

$\therefore \frac{dy}{y}= \frac{dx}{x}$

Now integrate..

**thomas_donald** · May 2nd 2010, 08:35 AM

Thank you very much.
I am able to solve x(t) and y(t) for System #2.
However, for System #1 if I integrate, I get y^2 = -(x^2)/2+x, and I am not sure how to get y(t) and x(t).

**kjchauhan** · May 2nd 2010, 08:44 PM

Originally Posted by kjchauhan

(1) Pl Note that,

$\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx}$

$\therefore \frac{dy}{dx}=\frac{(x-1)(y-1)}{-y(y-1)}$

$\therefore y \cdot dy= -(x-1) \cdot dx$

$\therefore y \cdot dy+(x-1) \cdot dx = 0$

Now integrate..

Same way,,,

(2)

$\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx}$

$\therefore \frac{dy}{dx}=\frac{y}{x)}$

$\therefore \frac{dy}{y}= \frac{dx}{x}$

Now integrate..

$y \cdot dy+(x-1) \cdot dx = 0$

$\therefore \frac{y^2}{2} + \frac{(x-1)^2}{2} = C$ (Constant)

**thomas_donald** · May 3rd 2010, 03:19 PM

Thank you very much.

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