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anonymous

  • 5 years ago

can anyone say how to solve a differential equation using method of undetermined coefficients??

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  1. anonymous
    • 5 years ago
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    http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

  2. anonymous
    • 5 years ago
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    give me the equation

  3. anonymous
    • 5 years ago
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    ok( D^2-5D+6)y=e^3x+sinx

  4. anonymous
    • 5 years ago
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    just tell me the assumed particular integral.....

  5. anonymous
    • 5 years ago
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    bad diff equation, what is D?

  6. anonymous
    • 5 years ago
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    D is the notation for operator d/dx....

  7. anonymous
    • 5 years ago
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    \[y^{\prime\prime}-5y^\prime+6y=\operatorname{e}^{3x}+\sin x\]

  8. anonymous
    • 5 years ago
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    yeah

  9. anonymous
    • 5 years ago
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    I think one of the best ways to solve this equation is by using Laplace transformation

  10. anonymous
    • 5 years ago
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    in the question it is specifically mentioned to use method of undetermined coefficients

  11. anonymous
    • 5 years ago
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    The particular solution will be the sum of particular solutions. Try Ae^(3x) + Bsin(x)

  12. anonymous
    • 5 years ago
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    solve for A and B

  13. anonymous
    • 5 years ago
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    no it will not work as e^3x is a term in complementary function

  14. anonymous
    • 5 years ago
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    particular solution \[y=Ae^{\lambda1 x}+Be^{\lambda2 x}\], lambda are complex numbers

  15. anonymous
    • 5 years ago
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    yeah...i just did it and it's a mess...too much to write out...buggin' out and going to bed...good luck.

  16. anonymous
    • 5 years ago
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    I thought the others were going to help you. Since e^(3x) is part of the homogeneous solution, try xAe^(3x) as one part. Since you have a trig. function too, add to this Msin(x) + Ncos(x). So try,\[y_p=Axe^{3x}+M \sin x +N \sin x\]

  17. anonymous
    • 5 years ago
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    I think I ended up with A=1, M=N=(1/10). Try it.

  18. anonymous
    • 5 years ago
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    \[C _{1} e^{2x} + C_{2}e^{3x} +0.1(\cos[x] + \sin[x])+xe^{3x}\]

  19. anonymous
    • 5 years ago
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    Yes

  20. anonymous
    • 5 years ago
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    thanx i ended up with the same solution guess the answer in the book was wrong

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