Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

MathSofiya Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    your y_c is wrong

    • one year ago
  2. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    it's -1

    • one year ago
  3. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    typo

    • one year ago
  4. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    mah bad...

    • one year ago
  5. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  6. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah r=-1

    • one year ago
  7. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    i was actually referring to the x by c2

    • one year ago
  8. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  9. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh I'm sorry I see it now

    • one year ago
  10. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    so do you know what y_p should be now?

    • one year ago
  11. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  12. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    are you using undetermined coefficients or variation of parameters>

    • one year ago
  13. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  14. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  15. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    please don't shoot me :S

    • one year ago
  16. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm willing to learn =)

    • one year ago
  17. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  18. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  19. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    oh that

    • one year ago
  20. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  21. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  22. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • one year ago
  23. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    does that make sense?

    • one year ago
  24. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    "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

    • one year ago
  25. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  26. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  27. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  28. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    sounds right

    • one year ago
  29. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    thank ya!

    • one year ago
  30. lgbasallote Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    welcome

    • one year ago
  31. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  32. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  33. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh I see

    • one year ago
  34. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok I'll try it again

    • one year ago
  35. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  36. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Hold up.

    • one year ago
  37. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  38. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    e-x?

    • one year ago
  39. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    e^(-x)?

    • one year ago
  40. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  41. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  42. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  43. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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'.

    • one year ago
  44. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  45. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  46. Algebraic! Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  47. MathSofiya Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah It kinda makes sense.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.