Linear System of Equations

Mar 2010
51
3
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.

\(\displaystyle u'\; =\; 4u\; -\; v\)
\(\displaystyle v'\; =\; -4u\; +\; 4v\)

Where v and u are dependent over x.
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
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.

\(\displaystyle u'\; =\; 4u\; -\; v\)
\(\displaystyle v'\; =\; -4u\; +\; 4v\)

Where v and u are dependent over x.
If you go to Chris' tutorial (the sticky at the top)

http://www.mathhelpforum.com/math-help/differential-equations/38182-differential-equations-tutorial.html

on page 12 you'll find what you need.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.

\(\displaystyle u'\; =\; 4u\; -\; v\)
\(\displaystyle v'\; =\; -4u\; +\; 4v\)

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.
 
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