# coupled equations

• November 8th 2009, 11:40 PM
Hikari Clover
coupled equations
http://i52.photobucket.com/albums/g37/mmmmms/aaa.jpg

dx/dt - 5x = -3Ae^2t

---->

2x-5x^2 = -3Ae^2t

what should i do to get the answer?

thx
• November 9th 2009, 11:04 PM
CaptainBlack
Quote:

Originally Posted by Hikari Clover
http://i52.photobucket.com/albums/g37/mmmmms/aaa.jpg

dx/dt - 5x = -3Ae^2t

---->

2x-5x^2 = -3Ae^2t

what should i do to get the answer?

thx

First this is a linear constant coefficeient non-homogeneous ODE. So the general solution is the sum of the general solution of the homogeneous problem:

$\frac{dx}{dt} - 5x = 0\ \ \ \ \ (1)$

and a particular integral of:

$
\frac{dx}{dt} - 5x = -3Ae^{2t}\ \ \ \ \ (2)
$

You should know how to find the general solution to $(1)$, and to find a particular integral of $(2)$ try a solution of the form:

$
x(t)=k e^{2t}
$

CB