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

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- Nov 8th 2009, 11:40 PMHikari Clovercoupled 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 - Nov 9th 2009, 11:04 PMCaptainBlack

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:

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

and a particular integral of:

$\displaystyle

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

$

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

$\displaystyle

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

$

CB