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melmel

  • 2 years ago

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

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  1. melmel
    • 2 years ago
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    |dw:1381226444353:dw|

  2. terenzreignz
    • 2 years ago
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    This WILL be messy... are you up for it? :D

  3. melmel
    • 2 years ago
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    yes

  4. terenzreignz
    • 2 years ago
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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?

  5. melmel
    • 2 years ago
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    \[m ^{^{2}}=11\]

  6. terenzreignz
    • 2 years ago
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    What does that mean? D:

  7. melmel
    • 2 years ago
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    \[y=e ^{a}\]

  8. terenzreignz
    • 2 years ago
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    Oh wait... there's no y' term?

  9. melmel
    • 2 years ago
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    get the higher order first D.E

  10. terenzreignz
    • 2 years ago
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    So basically, it's \[\Large y'' - 11y = xe^x\]?

  11. melmel
    • 2 years ago
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    \[m ^{2} +1 -12 =0\]

  12. melmel
    • 2 years ago
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    then m= square root of 11

  13. melmel
    • 2 years ago
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    then get the vlue of y

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

  15. melmel
    • 2 years ago
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    Im right? what is the next get the other y which from Xe^x

  16. melmel
    • 2 years ago
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    whic is y= Ae^u

  17. melmel
    • 2 years ago
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    take derivative until second derivative

  18. melmel
    • 2 years ago
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    ang get the value of A I stocked here

  19. melmel
    • 2 years ago
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    help guys if Im wrong :(

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

  21. melmel
    • 2 years ago
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    subst the value of Y" ,Y' and Y to the y′′+ y -12y

  22. melmel
    • 2 years ago
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    where did you get this y=Axex+Bex

  23. melmel
    • 2 years ago
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    @UnkleRhaukus

  24. UnkleRhaukus
    • 2 years ago
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    are you sure the second term in the question isn't a derivative?

  25. terenzreignz
    • 2 years ago
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    Odd, isn't it :D

  26. terenzreignz
    • 2 years ago
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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.

  27. UnkleRhaukus
    • 2 years ago
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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 }

  28. UnkleRhaukus
    • 2 years ago
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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 ?

  29. terenzreignz
    • 2 years ago
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    apparently offline :( Oh well, onwards and upwards XD

  30. terenzreignz
    • 2 years ago
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    ^_^ thanks

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