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melmel Group Title

linear D.E of nth order with constant coefficient...help guys .

  • one year ago
  • one year ago

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  1. melmel Group Title
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    |dw:1381226444353:dw|

    • one year ago
  2. terenzreignz Group Title
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    This WILL be messy... are you up for it? :D

    • one year ago
  3. melmel Group Title
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    yes

    • one year ago
  4. terenzreignz Group Title
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    For these kinds of differential equations, the general form of the solution is given by \[\Large f^{(h)}(x) + f^{(p)}(x)\] where the f^h is the general solution to the homogeneous equation while f^p is any particular solution of this equation. Can you find the general solution to the homogeneous equation?

    • one year ago
  5. melmel Group Title
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    \[m ^{^{2}}=11\]

    • one year ago
  6. terenzreignz Group Title
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    What does that mean? D:

    • one year ago
  7. melmel Group Title
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    \[y=e ^{a}\]

    • one year ago
  8. terenzreignz Group Title
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    Oh wait... there's no y' term?

    • one year ago
  9. melmel Group Title
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    get the higher order first D.E

    • one year ago
  10. terenzreignz Group Title
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    So basically, it's \[\Large y'' - 11y = xe^x\]?

    • one year ago
  11. melmel Group Title
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    \[m ^{2} +1 -12 =0\]

    • one year ago
  12. melmel Group Title
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    then m= square root of 11

    • one year ago
  13. melmel Group Title
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    then get the vlue of y

    • one year ago
  14. melmel Group Title
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    \[y=C{1}^{+\sqrt{11}x}+C{2}^{-\sqrt{11}x}\]

    • one year ago
  15. melmel Group Title
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    Im right? what is the next get the other y which from Xe^x

    • one year ago
  16. melmel Group Title
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    whic is y= Ae^u

    • one year ago
  17. melmel Group Title
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    take derivative until second derivative

    • one year ago
  18. melmel Group Title
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    ang get the value of A I stocked here

    • one year ago
  19. melmel Group Title
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    help guys if Im wrong :(

    • one year ago
  20. terenzreignz Group Title
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    Well, if it doesn't work with only one undetermined coefficient, did you try 2? Specifically this: \[\Large y = \color{blue}Axe^x + \color{red}Be^x\]

    • one year ago
  21. melmel Group Title
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    subst the value of Y" ,Y' and Y to the y′′+ y -12y

    • one year ago
  22. melmel Group Title
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    where did you get this y=Axex+Bex

    • one year ago
  23. melmel Group Title
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    @UnkleRhaukus

    • one year ago
  24. UnkleRhaukus Group Title
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    are you sure the second term in the question isn't a derivative?

    • one year ago
  25. terenzreignz Group Title
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    Odd, isn't it :D

    • one year ago
  26. terenzreignz Group Title
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    Oh by the way, I didn't actually "get" that Axe^x + Be^x, for all intents and purposes, it IS just a guess... see if it gets you the coefficients you want, since Axe^x alone doesn't seem to do it.

    • one year ago
  27. UnkleRhaukus Group Title
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    if not (yes strange), then i think you (@melmel ) got the complementary homogenous solution \[\large f^h(x)=y_c(x)=C_{1}e^{+\sqrt{11}x}+C_{2}e^{-\sqrt{11}x}\] {forgot to type the e's in the bases }

    • one year ago
  28. UnkleRhaukus Group Title
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    Which method to calculate the particular solution \[f^h(x)=y_p(x)\] are you going to use; undetermined coefficients, operator D , or variation of parameters, @melmel ?

    • one year ago
  29. terenzreignz Group Title
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    apparently offline :( Oh well, onwards and upwards XD

    • one year ago
  30. terenzreignz Group Title
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    ^_^ thanks

    • one year ago
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