# Thread: solve by systematic elimination...

1. ## solve by systematic elimination...

dx/dt = -5x - y

dy/dt = 4x-y

x(1) = 0 and y(1)= 1

ok so rewriting this problem....

Dx + 5x + y = 0

Dy - 4x + y = 0

rewriting it again I get...

x(D+5) + y = 0
-4x + y(D+1) = 0

so i chose to eliminate the x first and got my y to be...

y = C1e^-3t + C2te^-3t

and this is where i get confused, i thought i was doing the problem right but my book has a different answer. i know the initial values will replace the C1 and C2 values...it seems that the e^-3t should be e^-3t+3 which i dont see how they got...

any help would be appreciated..thanks in advance!

2. So your start seem good, but I do have to comment you really can put the x and y in front of the D terms. so we have

$\begin{array}{ccccc}
(D+5)x & + & y & =& 0 \\
-4x & + & (D+1)y & = & 0
\end{array}$

Now you want to eliminate x so we multiply the top equation by 4 and let $(D+5)$ act on the bottom equation to get

$\begin{array}{ccccc}
4(D+5)x & + & 4y & =& 0 \\
-4(D+5)x & + & (D+5)(D+1)y & = & 0
\end{array}$

Now if we add the two equations we get

$4y+(D+5)(D+1)y=0 \iff [4+(D+5)(D+1)]y=0$

$[D^2+6D+9]y=0 \iff (D+3)^2y=0$

This has complimentary solution

$y(t)=c_1e^{-3t}+c_2te^{-3t}$

Now if we plug this into the 2nd equation we get

$\displastyle x=\frac{y+y'}{4}$

$y'(t)=-3c_1e^{-3t}+c_2(e^{-3t}-3te^{-3t})$

This gives

$x(t)=\frac{1}{4}\left( -3c_1e^{-3t}+c_2(e^{-3t}-3te^{-3t})+c_1e^{-3t}+c_2te^{-3t}\right)$

$\displaystyle x(t)=\frac{1}{4}[(-2c_1+c_2)e^{-3t}-2c_2te^{-3t}]$

$0=(-2c_1+c_2)e^{-3}-c_2e^{-3} \iff -2c_1-c_2=0$

$1=c_1e^{-3}+c_2e^{-3} \iff e^{3}=c_1+c_2$

This gives $c_1=-e^{3} \quad c_2=2e^{3}$

If you put these into the two equations above you will get what you want

$y(t)=-e^{3}e^{-3t}+2e^{3}te^{-3t}=-e^{-3t+3}+2e^{-3t+3}$

3. ok let me go back and make my corrections..thanks alot!

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# solve de by systematic elimination

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