anonymous
  • anonymous
\[y''+2y'+y=xe^{-x}\] \[r^2+2r+1=0\] \[r_1=r_2=1\] \[y_c=c_1e^{-x}+c_2e^{-x}\] for y_p would I have (before I derive it and solve for A) \[y_p=Ae^{-x}\] or \[y_p=Axe^{-x}\] ?
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

lgbasallote
  • lgbasallote
your y_c is wrong
anonymous
  • anonymous
it's -1
anonymous
  • anonymous
typo

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
mah bad...
lgbasallote
  • lgbasallote
it should be \[y_c = c_1 e^{-x} + c_2 x e^{-x}\] see your mistake?
anonymous
  • anonymous
yeah r=-1
lgbasallote
  • lgbasallote
i was actually referring to the x by c2
lgbasallote
  • lgbasallote
if you have repeated roots you attach least power of x to the next arbitrary constant, remember?
anonymous
  • anonymous
oh I'm sorry I see it now
lgbasallote
  • lgbasallote
so do you know what y_p should be now?
anonymous
  • 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
  • lgbasallote
are you using undetermined coefficients or variation of parameters>
lgbasallote
  • lgbasallote
based on your question i presumed you were doing variation....which depends on the complementary solution....
anonymous
  • anonymous
I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]
anonymous
  • anonymous
please don't shoot me :S
anonymous
  • anonymous
I'm willing to learn =)
lgbasallote
  • lgbasallote
can you describe to me the method you're familiar with? does it involve deriving the yp?
anonymous
  • anonymous
according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS
lgbasallote
  • lgbasallote
oh that
anonymous
  • anonymous
yeah lol I guess THE METHOD OF VARIATION OF PARAMETERS is the next section....I haven't started that yet :P
lgbasallote
  • 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
  • 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
  • lgbasallote
does that make sense?
anonymous
  • 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
  • lgbasallote
yes. that. but in this case, we're doing y_p
anonymous
  • anonymous
\[y_p=Ae^{-x}+Bxe^{-x}\]
anonymous
  • anonymous
and then I just do the derivative twice to solve for A and B
lgbasallote
  • lgbasallote
sounds right
anonymous
  • anonymous
thank ya!
lgbasallote
  • lgbasallote
welcome
anonymous
  • 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
  • anonymous
y ' =-Ae^(-x) -(x-1)Be^(-x) you have an extra x in there..
anonymous
  • anonymous
oh I see
anonymous
  • anonymous
ok I'll try it again
anonymous
  • anonymous
I still get a zero for x....I'll show my work, give me a second to type it :)
anonymous
  • anonymous
Hold up.
anonymous
  • anonymous
I think you need a C*x^2*e-x term as part of your solution...
anonymous
  • anonymous
e-x?
anonymous
  • anonymous
e^(-x)?
anonymous
  • anonymous
Why a \[Cx^2e^{-x}\] ?
anonymous
  • anonymous
I think I'm gonna start a new thread. This one is getting too long
anonymous
  • anonymous
the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.
anonymous
  • 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
  • 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
  • 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
  • anonymous
Hope that helps; probably not, but I tried:)
anonymous
  • anonymous
Yeah It kinda makes sense.

Looking for something else?

Not the answer you are looking for? Search for more explanations.