Question

I'm dealing with a degenerate system of equations, such that
(D+2)x+(D+2)y=e^(-3t)
(D+3)x+(D+3)y=e^(-2t)

I've shown the degeneration of the system using the operational determinant, but am at a loss as to where to go from here.

Answer 1

@amistre64 Tada (that's a neat little feature I didn't know about)

Answer 2

assuming your trying to find x and y; id cramer it up

Answer 3

and youll have to define the terminology since I got no idea what you said :)

Answer 4

Ok, are you familiar with wronkskians?

Answer 5

Wrongskins are cramers

\begin{pmatrix}(D+2)x+(D+2)y=e^(-3t)
(D+3)x+(D+3)y=e^(-2t
\end{pmatrix}

Answer 6

ok,
What I've done so far is proven that the equation either has 0 or \[\infty\] solutions. (that's the operational determinant stuff).
My next step is to figure out which of the two options it is.

Answer 7

What I've tried is setting the two equations equal to each other by multiplying by the respective 'e's and then solving. The part I'm a bit fuzzy on is whether I can factor out D like I would a normal constant/variable

Answer 8

\[\begin{pmatrix}
(D+2)&(D+2)&|&e^{-3t} \\
(D+3)&(D+3)&|&e^{-2t}
\end{pmatrix}\]

\[\begin{pmatrix}
(D+2)&(D+2)\\
(D+3)&(D+3)
\end{pmatrix}
, \begin{pmatrix}
(D+2)&e^{-3t} \\
(D+3)&e^{-2t}
\end{pmatrix},
\begin{pmatrix}
e^{-3t}&(D+2) \\
e^{-2t}&(D+3)
\end{pmatrix}\]

Answer 9

D, in this case, represents....well something I don't remember the name of. But effectively, D means "the derivative of" as best as I can grok it.

Answer 10

write, ive seen the D notation before, just havent used it alot

Answer 11

it's fairly new to me, unfortunately, so I haven't a firm grasp on it yet.

Answer 12

What did this look like before you dabbled your toes in it? :)

why this thing freezes from time to time is beyond me

Answer 13

That is actually the exact information I was given. They wanted me to prove it's degenerativness (the 0 or infinite part) and then determine if it has 0 or infinite solutions by attempting to solve it.

Answer 14

http://www.ma.hw.ac.uk/~simonm/linalg.pdf

page 31 is talking about degenerates ... im reading up on it if you wanna see if that pertains to your case

Answer 15

what I did was:
(d+2) = L1, (d+3) = L2

then I multiplied the first equation by L2 and the second by L1, getting
L1L2x+L2L1y = L2e^(-3t)
L1L2x+L2L1y = L1e^(-2t)

I set L2e^(-3t) = L1e^(-2t)
then expanded back to
(d+3)e^(-3t) = (d+2)e^(-2t)
distributed,
d*e^(-3t) + 3*e^(-3t) = d*e^(-2t) + 2*e^(-2t)
put the 'd's on one side, and the rest on the other, then derived.
d*e^(-3t) - d*e^(-2t) = -3*e^(-3t) +2*e^(-2t)
-3*e^(-3t) +2*e^(-2t) = -3*e^(-3t) +2*e^(-2t)

because they are equal, I deduced that there are infinite solutions to the system.

Answer 16

lol, im still working on definitions :)

If a real root is repeated, the solutions are degenerate and the linearly independent solutions are modified: \(y=c_1e^{rx}+c_2xe^{rx}+c_3x^2e^{rx}...\)

Answer 17

im not sure how to read the L1 and d+2 and such that you have; its a bit too abstract for me

Answer 18

Ah, well going over it a second time makes me think I'm at least close if not straight-up correct. Thanks for taking the time to look over it

Answer 19

your welcome, and good luck with it :)

Answer 20

thanks!