Question

\[y''+2y'+y=xe^{-x}\]
\[r^2+2r+1=0\]
\[r_1=r_2=1\]
\[y_c=c_1e^{-x}+c_2e^{-x}\]
for y_p would I have (before I derive it and solve for A)
\[y_p=Ae^{-x}\]
or
\[y_p=Axe^{-x}\]
?

Answer 1

your y_c is wrong

Answer 2

it's -1

Answer 3

typo

Answer 4

mah bad...

Answer 5

it should be \[y_c = c_1 e^{-x} + c_2 x e^{-x}\]
see your mistake?

Answer 6

yeah r=-1

Answer 7

i was actually referring to the x by c2

Answer 8

if you have repeated roots you attach least power of x to the next arbitrary constant, remember?

Answer 9

oh I'm sorry I see it now

Answer 10

so do you know what y_p should be now?

Answer 11

what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?

Answer 12

are you using undetermined coefficients or variation of parameters>

Answer 13

based on your question i presumed you were doing variation....which depends on the complementary solution....

Answer 14

I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]

Answer 15

please don't shoot me :S

Answer 16

I'm willing to learn =)

Answer 17

can you describe to me the method you're familiar with?

does it involve deriving the yp?

Answer 18

according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS

Answer 19

oh that

Answer 20

yeah lol I guess
THE METHOD OF VARIATION OF PARAMETERS
is the next section....I haven't started that yet :P

Answer 21

\[y_p = Ae^{bx}\] only works if you don't have repeated roots. for repeated roots, you use \[y_p = Ae^{bx} + Bxe^{bx}\]

Answer 22

because like i said before, if you have repeated roots, you attach the least power of x to the succeeding constant. Since you have xe^-x on the right side, there must be an e^-x before it

Answer 23

does that make sense?

Answer 24

"least power of x to the succeeding constant" meaning
\[y_c = c_1 e^{-x} + c_2 x e^{-x}\]
referring to the x after the c_2

Answer 25

yes. that. but in this case, we're doing y_p

Answer 26

\[y_p=Ae^{-x}+Bxe^{-x}\]

Answer 27

and then I just do the derivative twice to solve for A and B

Answer 28

sounds right

Answer 29

thank ya!

Answer 30

welcome

Answer 31

What did I do wrong? I get 0=x
\[y_p=Ae^{-x}+Bxe^{-x}\]
\[y_p'=-Ae^{-x}-Bxe^{-x}(x-1)\]
\[y_p''=Ae^{-x}+Be^{-x}(x-2)\]

Answer 32

y ' =-Ae^(-x) -(x-1)Be^(-x)

you have an extra x in there..

Answer 33

oh I see

Answer 34

ok I'll try it again

Answer 35

I still get a zero for x....I'll show my work, give me a second to type it :)

Answer 36

Hold up.

Answer 37

I think you need a C*x^2*e-x term as part of your solution...

Answer 38

e-x?

Answer 39

e^(-x)?

Answer 40

Why a
\[Cx^2e^{-x}\]
?

Answer 41

I think I'm gonna start a new thread. This one is getting too long

Answer 42

the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.

Answer 43

so without a term that doesn't end up being zero, you're not going to be able to ever get your DE to be 'forced'.

Answer 44

There's a quantitative way to justify it, but I can't recall how it goes. The point is xe^-x is the forcing term and one of the solutions to the homogeneous. So you need another term in your yp if you're ever going to get an actual solution that satisfies the original non-homogeneous equation.

Answer 45

Qualitatively speaking, you need a faster transient, because the forcing term is resonant with the system, so the whole thing is getting driven to x*e^-x harder than any of the terms you're using now can drive it.

Answer 46

Hope that helps; probably not, but I tried:)

Answer 47

Yeah It kinda makes sense.