# Linear System of Equations

• May 15th 2010, 04:47 PM
jameselmore91
Linear System of Equations
I'm having some trouble with Linear Systems of Equations, I get the basic concept but it would be helpful if someone could do the following problem step by step.

$u'\; =\; 4u\; -\; v$
$v'\; =\; -4u\; +\; 4v$

Where v and u are dependent over x.
• May 16th 2010, 05:44 AM
Jester
Quote:

Originally Posted by jameselmore91
I'm having some trouble with Linear Systems of Equations, I get the basic concept but it would be helpful if someone could do the following problem step by step.

$u'\; =\; 4u\; -\; v$
$v'\; =\; -4u\; +\; 4v$

Where v and u are dependent over x.

If you go to Chris' tutorial (the sticky at the top)

http://www.mathhelpforum.com/math-he...-tutorial.html

on page 12 you'll find what you need.
• May 16th 2010, 06:38 AM
HallsofIvy
Quote:

Originally Posted by jameselmore91
I'm having some trouble with Linear Systems of Equations, I get the basic concept but it would be helpful if someone could do the following problem step by step.

$u'\; =\; 4u\; -\; v$
$v'\; =\; -4u\; +\; 4v$

Where v and u are dependent over x.

The most fundamental way to do this is to differentiate the first equation again gettin u''= 4u'- v'. From the second equation, v'= -4u+ 4v so we can replace it: u''= 4u'- (-4u+ 4v)= 4u'+ 4u- 4v. We still have a "v" to eliiminate but from the first equation, v= 4u- u' so that becomes u''= 4u'+ 4u- 4(4u- u')= 8u' -12u. that is, u''- 8u+ 12u= 0. Solve that equation for the general solution, u, then use v= 4u- u' to find v.