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

- anonymous

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- lgbasallote

your y_c is wrong

- anonymous

it's -1

- anonymous

typo

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## More answers

- anonymous

mah bad...

- lgbasallote

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

- anonymous

yeah r=-1

- lgbasallote

i was actually referring to the x by c2

- lgbasallote

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

- anonymous

oh I'm sorry I see it now

- lgbasallote

so do you know what y_p should be now?

- anonymous

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

- lgbasallote

are you using undetermined coefficients or variation of parameters>

- lgbasallote

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

- anonymous

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

- anonymous

please don't shoot me :S

- anonymous

I'm willing to learn =)

- lgbasallote

can you describe to me the method you're familiar with?
does it involve deriving the yp?

- anonymous

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

- lgbasallote

oh that

- anonymous

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

- lgbasallote

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

- lgbasallote

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

- lgbasallote

does that make sense?

- anonymous

"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

- lgbasallote

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

- anonymous

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

- anonymous

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

- lgbasallote

sounds right

- anonymous

thank ya!

- lgbasallote

welcome

- anonymous

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

- anonymous

y ' =-Ae^(-x) -(x-1)Be^(-x)
you have an extra x in there..

- anonymous

oh I see

- anonymous

ok I'll try it again

- anonymous

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

- anonymous

Hold up.

- anonymous

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

- anonymous

e-x?

- anonymous

e^(-x)?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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'.

- anonymous

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.

- anonymous

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.

- anonymous

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

- anonymous

Yeah It kinda makes sense.

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