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MathSofiya

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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    your y_c is wrong

  2. MathSofiya
    • 2 years ago
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    it's -1

  3. MathSofiya
    • 2 years ago
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    typo

  4. MathSofiya
    • 2 years ago
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    mah bad...

  5. lgbasallote
    • 2 years ago
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    it should be \[y_c = c_1 e^{-x} + c_2 x e^{-x}\] see your mistake?

  6. MathSofiya
    • 2 years ago
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    yeah r=-1

  7. lgbasallote
    • 2 years ago
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    i was actually referring to the x by c2

  8. lgbasallote
    • 2 years ago
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    if you have repeated roots you attach least power of x to the next arbitrary constant, remember?

  9. MathSofiya
    • 2 years ago
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    oh I'm sorry I see it now

  10. lgbasallote
    • 2 years ago
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    so do you know what y_p should be now?

  11. MathSofiya
    • 2 years ago
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    what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?

  12. lgbasallote
    • 2 years ago
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    are you using undetermined coefficients or variation of parameters>

  13. lgbasallote
    • 2 years ago
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    based on your question i presumed you were doing variation....which depends on the complementary solution....

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

  15. MathSofiya
    • 2 years ago
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    please don't shoot me :S

  16. MathSofiya
    • 2 years ago
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    I'm willing to learn =)

  17. lgbasallote
    • 2 years ago
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    can you describe to me the method you're familiar with? does it involve deriving the yp?

  18. MathSofiya
    • 2 years ago
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    according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS

  19. lgbasallote
    • 2 years ago
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    oh that

  20. MathSofiya
    • 2 years ago
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    yeah lol I guess THE METHOD OF VARIATION OF PARAMETERS is the next section....I haven't started that yet :P

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

  22. lgbasallote
    • 2 years ago
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    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

  23. lgbasallote
    • 2 years ago
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    does that make sense?

  24. MathSofiya
    • 2 years ago
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    "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

  25. lgbasallote
    • 2 years ago
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    yes. that. but in this case, we're doing y_p

  26. MathSofiya
    • 2 years ago
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    \[y_p=Ae^{-x}+Bxe^{-x}\]

  27. MathSofiya
    • 2 years ago
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    and then I just do the derivative twice to solve for A and B

  28. lgbasallote
    • 2 years ago
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    sounds right

  29. MathSofiya
    • 2 years ago
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    thank ya!

  30. lgbasallote
    • 2 years ago
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    welcome

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

  32. Algebraic!
    • 2 years ago
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    y ' =-Ae^(-x) -(x-1)Be^(-x) you have an extra x in there..

  33. MathSofiya
    • 2 years ago
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    oh I see

  34. MathSofiya
    • 2 years ago
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    ok I'll try it again

  35. MathSofiya
    • 2 years ago
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    I still get a zero for x....I'll show my work, give me a second to type it :)

  36. Algebraic!
    • 2 years ago
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    Hold up.

  37. Algebraic!
    • 2 years ago
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    I think you need a C*x^2*e-x term as part of your solution...

  38. MathSofiya
    • 2 years ago
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    e-x?

  39. MathSofiya
    • 2 years ago
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    e^(-x)?

  40. MathSofiya
    • 2 years ago
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    Why a \[Cx^2e^{-x}\] ?

  41. MathSofiya
    • 2 years ago
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    I think I'm gonna start a new thread. This one is getting too long

  42. Algebraic!
    • 2 years ago
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    the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.

  43. Algebraic!
    • 2 years ago
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    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'.

  44. Algebraic!
    • 2 years ago
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    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.

  45. Algebraic!
    • 2 years ago
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    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.

  46. Algebraic!
    • 2 years ago
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    Hope that helps; probably not, but I tried:)

  47. MathSofiya
    • 2 years ago
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    Yeah It kinda makes sense.

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