# doomsday model example

• Jun 5th 2010, 07:01 PM
yen yen
doomsday model example
i was looking up for questions regarding this topic and came across a question.

dx/dt=3x-2

i think to get the equilibrium solution you set x(0)= x_0 and get
x=2/3 + (x_0 - 2/3)*e^3t
*

could anyone tell me how to get that? i tried integrating but it doesn't work
• Jun 5th 2010, 07:12 PM
Prove It
Quote:

Originally Posted by yen yen
i was looking up for questions regarding this topic and came across a question.

dx/dt=3x-2

i think to get the equilibrium solution you set x(0)= x_0 and get
x=2/3 + (x_0 - 2/3)*e^3t
*

could anyone tell me how to get that? i tried integrating but it doesn't work

$\frac{dx}{dt} = 3x - 2$

$\frac{1}{3x - 2}\,\frac{dx}{dt} = 1$

$\int{\frac{1}{3x - 2}\,\frac{dx}{dt}\,dt} = \int{1\,dt}$

$\int{\frac{1}{3x - 2}\,dx} = t + C_1$

$\frac{1}{3}\ln{|3x - 2|} + C_2 = t + C_1$

$\frac{1}{3}\ln{|3x - 2|} = t + C$ where $C = C_1 - C_2$

$\ln{|3x - 2|} = 3t + 3C$

$|3x - 2| = e^{3t + 3C}$

$|3x - 2| = e^{3C}e^{3t}$

$3x - 2 = \pm e^{3C}e^{3t}$

$3x - 2 = Ae^{3t}$ where $A = \pm e^{3C}$

$3x = Ae^{3t} + 2$

$x = Be^{3t} + \frac{2}{3}$ where $B = \frac{1}{3}A$.

Now if you let $x(0) = x_0$, we can solve for $B$...

$x_0 = Be^{3(0)} + \frac{2}{3}$

$x_0 = B + \frac{2}{3}$

$x_0 - \frac{2}{3} = B$.

Therefore

$x = \left(x_0 - \frac{2}{3}\right)e^{3t} + \frac{2}{3}$.
• Jun 6th 2010, 05:35 PM
yen yen
thanks. i forgot about seperable d.e
• Jun 7th 2010, 02:00 AM
Ackbeet
Whether or not a d.e. is separable should be one of the first things you look for, in my opinion. As Griffiths says in his Quantum Mechanics textbook, "In that case the Schroedinger equation can be solved by the method of separation of variables (the physicist's first line of attack on any partial differential equation):..." - p. 20. I would argue that should also be true of the ODE.