I have a few questions about finding the general solution of first-order linear differential equations..

please help, I have no clue!! =/

http://i28.tinypic.com/2yv3343.gif

http://i31.tinypic.com/2818w02.gif

http://i28.tinypic.com/n81ee.gif

Printable View

- Apr 19th 2008, 10:10 PMiloveyou33linear differential equations??
I have a few questions about finding the general solution of first-order linear differential equations..

please help, I have no clue!! =/

http://i28.tinypic.com/2yv3343.gif

http://i31.tinypic.com/2818w02.gif

http://i28.tinypic.com/n81ee.gif - Apr 19th 2008, 10:14 PMJhevon
the integrating factor method takes care of all these. see post #21 here

- Apr 19th 2008, 10:20 PMiloveyou33
it doesn't make sense though..

is there a simpler way of explaining it?

I don't understand what "the integrating factor" is in the first place.. - Apr 19th 2008, 10:33 PMJhevon
it is just something you multiply through by to make the left hand side the derivative given by the product and the integrating factor. for the first one:

$\displaystyle y' + 2y = 3e^t$

(with practice, you will be able to recognize the integrating factor immediately, but let's go through the method to see it)

the integrating factor is $\displaystyle e^{\int 2~dt} = e^{2t}$

multiply through by the integrating factor, we get:

$\displaystyle e^{2t}y' + 2e^{2t}y = 3e^{3t}$

now the left hand side is the result of differentiating $\displaystyle e^{2t}y$ by the product rule, thus

$\displaystyle (e^{2t}y)' = 3e^{3t}$

integrate both sides, we get:

$\displaystyle e^{2t}y = e^{3t} + C$

$\displaystyle \Rightarrow y = e^t + Ce^{-2t}$

the others are done similarly - Apr 20th 2008, 08:51 AMKrizalid
- Apr 20th 2008, 09:44 AMJhevon