Question

http://tinypic.com/r/29g2amx/8
Difficult question using elimination method for a given linear system and an inverse Laplace transform

Answer 1

\[\begin{cases}
x''+y'=x+y+\dfrac{1}{2}e^{2t}&(1)\\\\
x'+y'=e^{2t}&(2)
\end{cases}\]
Try differentiating both equations:
\[\begin{cases}x'''+y''=x'+y'+e^{2t}&(3)\\
x''+y''=2e^{2t}&(4)
\end{cases}\]
Substitute (2) and (4) into (3), and you'll have a linear ODE in \(x\).

Answer 2

Partial fractions will work nicely for the inverse transform, but the difficulty is finding the roots to the denominator polynomial. Seeing as this is a take-home, though, it seems you have any method at your disposal. I suggest a calculator.

You'll find that
\[s^3+4s^2-7s+30=(s+6)(s^2-2s+5)\]
As for how you'd arrive at these factors by hand, I suppose you could take factors of 30, call them \(k\), and divide the given cubic by \(x-k\)... This will work once you set \(k=-6\), but that would involve some trial and error.

You could consider giving this method a try, too: http://www.sosmath.com/algebra/factor/fac11/fac11.html

Answer 3

Oh hmm. I use a slightly different method.
\[(D^2-1)x (D-1)y = 1/2 e^(2t)\]
\[(D-1)[Dx+Dy=e^(2t)]\]
\[(D^3-D)x-D(D-1)y \]=
\[(D^3-D)x+D(D-1)y\]
\[(-D^3+D^2)x=0\]
then you turn all that into
\[-x'''+x"=0\]
\[R^3+R^2=0\]
and solve for R
and then for some reason my mind just blanks on the next part. Which is the easiest part too...

Answer 4

For #2 I did this
\[g(t+2)=f(t)\]
\[g(t)=f(t-2)\]
\[L \left\{ u(t-a)*g(t) \right\} = e ^{-as} L \left\{ g(t+2) \right\}\]
\[x'+1/2 e ^{2t}+c-x=1/2e ^{2t}+ct+k\]
\[x'-x=ct+\omega\]
then I get stuck on how to finish....
Help me please!

Answer 5

I don't follow your work for #2. Are you sure it's the same problem?

Answer 6

Oh oops, that was a continuation of #1.

Answer 7

Is it on the right track?

Answer 8

is wat on track

Answer 9

I'd made a mistake the first time around, your final equation in \(x\) is fine. You went about it somewhat differently, but we get the same result. Sorry.
\[\begin{cases}
x''+y'=x+y+\dfrac{1}{2}e^{2t}&(1)\\\\
x'+y'=e^{2t}&(2)
\end{cases}\]
is equivalent to
\[\begin{cases}
D^2x+Dy=x+y+\dfrac{1}{2}e^{2t}&(1')\\\\
Dx+Dy=e^{2t}&(2')
\end{cases}\]
Differentiating (1') and (2') is the same as distributing \(D\) over both equations:
\[\begin{cases}
D^3x+D^2y=Dx+Dy+e^{2t}&(3)\\\\
D^2x+D^2y=2e^{2t}&(4)
\end{cases}\]
Substitute (2') into (3) and you get
\[\begin{cases}
D^3x+D^2y=e^{2t}+e^{2t}=2e^{2t}&(5)\\\\
D^2x+D^2y=2e^{2t}&(4)
\end{cases}\]
Eliminating \(D^2y\) gives
\[D^3x-D^2x=0\]
Like you said, this gives the characteristic equation
\[r^3-r=0~~\implies~~r(r^2-1)=0~~\implies~~r=0,\pm1\]
and so your solution for \(x\) takes the form
\[x(t)=C_1+C_2e^t+C_3e^{-t}\]