first order systems of equations--- HELP!

friesepferd

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Jun 27, 2006
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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!
 
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...
 
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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