Question

undetermined coefficients...
y"-y'+2y=-6xe^-x

Answer 1

i know how to do make y = D
\[D^2-D-2 = 0\] then \[(D+1)(D-2) \] there for it becomes \[C1e^-1x + C2e^2x\]

Answer 2

for the general solution

Answer 3

particular solution first. \[\huge Axe^{-x}\]

Answer 4

wher did Axe^-x come from?

Answer 5

multiple roots {-1}

Answer 6

what?

Answer 7

my mistake. the particular solution is \[\large Axe^{-x} + Bx^2e^{-x}\]now you're really confused.

Answer 8

yes i am can you show me how that happened?

Answer 9

let's begin with this. do you know ho to write a homogeneous DE for a particular solution, say \[\large y=5e^{3x}\]

Answer 10

um.. not really 100% sure

Answer 11

let's do it backwards then. you concluded that the compl. solution to \[\large (D+1)(D-2)y = 0 \] is \[\large c_1e^{-x} + c_2e^{2x}\] is based on

Answer 12

is based on what result?

Answer 13

general solution?

Answer 14

how did you get \[\huge e^{-x}\] and \[\huge e^{2x}\]? specifically, why did you choose -1 and +2 as coefficients of x?

Answer 15

i found out that it was \[D^2-D-2 \] then factor it out \[(D+1)(D-2) \] which becomes D = -1, 2

Answer 16

we build our solution from that. \[\large e^{-x}\] for (D+1) and \[\large e^{2x}\] for (D-2)
so for \[\large c_1e^{5x} + c_2e^{4x} + c_3e^{-2x} \] the polynomial in D is \[(D-5)(D-4)(D+2)\] agreed?

Answer 17

what i dont understand is how did we get e^5x and e^4x?

Answer 18

\[\large (D-5)\] creates the solution \[\large ce^{5x}\]
the coefficient of x is the root of (D-5). see the connection?

Answer 19

yes i know that d-5 is ce^5x but where did d-5 come from that is the question

Answer 20

its not where it came from. when you see a (D-5) in a problem, you create \[\large ce^{5x}\] as part of the compl. solution to that problem where there is some (D-5)

Answer 21

oh i see! thank you so much now i understand! THANKS for hanging with me!

Answer 22

are you still game?

Answer 23

yea i actually understood it danka!

Answer 24

if you see (D-5)(D-5) in some problem, what functions do you create as part of the compl. solution?

Answer 25

(D-5)^2 are you asking about that?

Answer 26

yes.