first order systems of equations--- HELP!

friesepferd

New member
Joined
Jun 27, 2006
Messages
1
how would you do a question such as the following:

rewrite the given system as a first order system. also indicate what quantities must be specified in an appropriate set of auxiliary conditions for the given system at a point t0



x'1 - x'2 - x1 = cos t
x'2 - 3x1 = e^t


the answer is the following. i just dont know how to get there:

x'1 = 4x1 + cos t + e^t x1(t0)=k1
x'2 = 3x1 +e^t x2(t0)=k2


please help!. my teacher is out of the country and we are suppose to learn this stuff from the book but the book sux and this problem as well as these next two are due real soon!:

x'1 - x'2 = e^t
x''2 - x1 - x2 - x'2 = sin t


and the last problem:

x'''1 - x'1 +x1*x'2 = sin t
x''1 - x2*x'1 - x''2 = cos t


thanks!
 

daon

Senior Member
Joined
Jan 27, 2006
Messages
1,284
x'1 - x'2 - x1 = cos t
x'2 - 3x1 = e^t

\(\displaystyle \L \left[ \begin{array}{ccc}
1 & -1 & -1 & : & cost\\
0 & 1 & -3 & : & e^t \end{array} \right] \\
R_1 \leftarrow R_1 + R_2 \Rightarrow \\
\left[ \begin{array}{ccc}
1 & 0 & -4 & : & cost + e^t\\
0 & 1 & -3 & : & e^t \end{array} \right] \\\)

Then \(\displaystyle x_1' - 4x_1 = cost + e^t\)
and \(\displaystyle x_2' -3x_1 = e^t\)
...which is the same thing as your answer.

I'm not sure what you mean by x1(t0) = k1...
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,587
Hello, friesepferd!

Those subscripts are annoying . . . I'll use \(\displaystyle x\) and \(\displaystyle y.\)

Rewrite the given system as a first order system.
Also indicate what quantities must be specified in an appropriate set of auxiliary conditions
for the given system at a point \(\displaystyle t_o.\)

\(\displaystyle \,[1]\;\;x'\, - \,y'\, -\, x \:=\:\cos t\)
\(\displaystyle [2]\;\;\;\;\;y'\, -\, 3x \:=\: e^t\)


The answer is the following. I just don't know how to get there \(\displaystyle \;\) . . . you don't?

\(\displaystyle x' \:= \:4x\,+\,\cos t\,+\,e^t\;\;\;x(t_o)\,=\,k_1\)
\(\displaystyle y'\: = \:3x\,+\,e^t\;\;\;y(t_o)\,=\,k_2\)
From [2], we have: \(\displaystyle \,y'\:=\:3x\,+\,e^t\;\) . . . wow!

Substitute into [1]: \(\displaystyle \,x'\,-\,(3x\,+\,e^t)\,-\,x\;=\;\cos t\;\;\Rightarrow\;\;x'\:=\:4x\,+\,\cos t\,+\,e^t\;\) . . . golly!
 
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