Question

Ok...we may need to use a linear differential operator on this equation (D-2)^2(D-1)y=4e^(2x)
I know the Yh is D = 2 D =1
so c1e^2x+c2xe^2x+c3e^x
the Yp is Ax^2e^2x because Ae^2x and Axe^2x doesn't work...

Answer 1

so now there's two things going on... we may need to set the e^2x aside and take the derivative of Yp

Answer 2

@OOOPS

Answer 3

the answer is supposed to be 2 but how can I get there??????

Answer 4

A is supposed to be 2 so that Yp=2x^2e^2x

Answer 5

so set A aside and product rule x^2e^2x

Answer 6

well I set the e^2x aside because supposedly that disappears.

Answer 7

@dan815 we're at this again rofl

Answer 8

it's one of those linear operator questions. x)

Answer 9

@raffle_snaffle

Answer 10

we could use this for the problem
http://math.wallawalla.edu/~duncjo/courses/math312/spring07/notes/4-5_math312.pdf

Answer 11

yeah I have just one more of those

Answer 12

let me just find it seriously I'm tired of dealing with that time consuming crud. and i have 2 exams to study for. so this is just eating me up

Answer 13

(D-1)(D^2-2D+2)y=e^xcosx

Answer 14

just that and that's it for that stuff. -_- it's worse than variation of parameters.

Answer 15

so Yh for (D-1)(D^2-2D+2)y=e^xcosx
c1e^x+c2e^(x)cosx+c3e^(x)sinx

Answer 16

and then since e^xcosx is already on there... we can't do e^xcosx+e^xsinx for Yp...should be xe^(x)cosx+xe^(x)sinx=Yp

Answer 17

I think finding the guess is easy...doing it is oy

Answer 18

crud I forgot the letters.

Answer 19

but prof didn't do annihilation well can you teach me ?

Answer 20

oh so how does Axcosxe^x+Bxsinxe^x ...alright then we take derivatives...at least three of them x.x

Answer 21

do I just set the A and B aside and take derivatives

Answer 22

ok so what's the plan?