Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

your y_c is wrong
it's -1
typo

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

mah bad...
it should be \[y_c = c_1 e^{-x} + c_2 x e^{-x}\] see your mistake?
yeah r=-1
i was actually referring to the x by c2
if you have repeated roots you attach least power of x to the next arbitrary constant, remember?
oh I'm sorry I see it now
so do you know what y_p should be now?
what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?
are you using undetermined coefficients or variation of parameters>
based on your question i presumed you were doing variation....which depends on the complementary solution....
I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]
please don't shoot me :S
I'm willing to learn =)
can you describe to me the method you're familiar with? does it involve deriving the yp?
according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS
oh that
yeah lol I guess THE METHOD OF VARIATION OF PARAMETERS is the next section....I haven't started that yet :P
\[y_p = Ae^{bx}\] only works if you don't have repeated roots. for repeated roots, you use \[y_p = Ae^{bx} + Bxe^{bx}\]
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
does that make sense?
"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
yes. that. but in this case, we're doing y_p
\[y_p=Ae^{-x}+Bxe^{-x}\]
and then I just do the derivative twice to solve for A and B
sounds right
thank ya!
welcome
What did I do wrong? I get 0=x \[y_p=Ae^{-x}+Bxe^{-x}\] \[y_p'=-Ae^{-x}-Bxe^{-x}(x-1)\] \[y_p''=Ae^{-x}+Be^{-x}(x-2)\]
y ' =-Ae^(-x) -(x-1)Be^(-x) you have an extra x in there..
oh I see
ok I'll try it again
I still get a zero for x....I'll show my work, give me a second to type it :)
Hold up.
I think you need a C*x^2*e-x term as part of your solution...
e-x?
e^(-x)?
Why a \[Cx^2e^{-x}\] ?
I think I'm gonna start a new thread. This one is getting too long
the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.
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'.
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.
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.
Hope that helps; probably not, but I tried:)
Yeah It kinda makes sense.

Not the answer you are looking for?

Search for more explanations.

Ask your own question