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lilMissMindset Group Title

DIFFERENTIAL EQUATION: help please, anybody. (D^2 +4)y =4sin^2 (x)

  • one year ago
  • one year ago

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  1. shubhamsrg Group Title
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    y= a cos2x ?

    • one year ago
  2. shubhamsrg Group Title
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    guess that substn should work..

    • one year ago
  3. lilMissMindset Group Title
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    aaa. i think i got it. wait, i'll try.

    • one year ago
  4. shubhamsrg Group Title
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    ohh wait,,i didnt see a "y" also in there. !!

    • one year ago
  5. Algebraic! Group Title
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    @shubhamsrg I think it's just written in 'operator' notation... \[\frac{ d^2y }{dt^2 } +4y = 4\sin^2x\] correct @lilMissMindset ?

    • one year ago
  6. Algebraic! Group Title
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    heh.

    • one year ago
  7. shubhamsrg Group Title
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    there's a y also with D^2 .. i had got that,,i didnt see a y being multiplied to whole that time..sorry..

    • one year ago
  8. Algebraic! Group Title
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    \[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\] rather.

    • one year ago
  9. lilMissMindset Group Title
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    yes, tat's it

    • one year ago
  10. lilMissMindset Group Title
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    that*

    • one year ago
  11. Algebraic! Group Title
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    yes he's correct I believe y =a*cos^2 x

    • one year ago
  12. Algebraic! Group Title
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    did you get it to work?

    • one year ago
  13. Algebraic! Group Title
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    that's the particular soln. anyway...

    • one year ago
  14. lilMissMindset Group Title
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    isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx

    • one year ago
  15. Algebraic! Group Title
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    why?

    • one year ago
  16. lilMissMindset Group Title
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    since sinx is twice? applying case 4 of undetermined coefficients

    • one year ago
  17. lilMissMindset Group Title
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    where it has some repeated imaginary root

    • one year ago
  18. Algebraic! Group Title
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    look d^2 (a*cos^2(x)) /dx^2 +4a cos^2(x) = 4sin^2(x) second derivative of a*cos^2(x) is 2a*cos^2(x) - 2a*sin^2(x)

    • one year ago
  19. Algebraic! Group Title
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    this is for the particular solution...

    • one year ago
  20. Algebraic! Group Title
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    finding the homogeneous solution shouldn't be a problem...

    • one year ago
  21. Algebraic! Group Title
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    \[\lambda ^2 +4 =0\]

    • one year ago
  22. Algebraic! Group Title
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    lambda = 2i

    • one year ago
  23. Algebraic! Group Title
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    ok?

    • one year ago
  24. lilMissMindset Group Title
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    i'll work on it. please guide me still, in case i get stuck.

    • one year ago
  25. Algebraic! Group Title
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    hold on... can you tell me what part is giving you problems?

    • one year ago
  26. lilMissMindset Group Title
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    getting Yp gives me a lot of trouble. since you told me that it's acos^2 x, i am differentiating it now.

    • one year ago
  27. Algebraic! Group Title
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    ok:)

    • one year ago
  28. lilMissMindset Group Title
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    how about if the given is cos^x, then Yp=asin^2 x?

    • one year ago
  29. Algebraic! Group Title
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    if the forcing term is cos^2 x you mean?

    • one year ago
  30. lilMissMindset Group Title
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    yes.

    • one year ago
  31. lilMissMindset Group Title
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    i forgot 2, lol.

    • one year ago
  32. Algebraic! Group Title
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    yeah, for the same equation, I believe that a*sin^2 x would be the trial solution then...

    • one year ago
  33. Algebraic! Group Title
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    yes. it would. just checked it.

    • one year ago
  34. lilMissMindset Group Title
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    thanks!

    • one year ago
  35. Algebraic! Group Title
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    sure:)

    • one year ago
  36. lilMissMindset Group Title
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    darn, i can't do it. show me a brief solution please. so i can learn how.

    • one year ago
  37. lilMissMindset Group Title
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    just for the Yp part. swear.

    • one year ago
  38. Algebraic! Group Title
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    well, I did it already (a*cos^2(x))" +4a cos^2(x) = 4sin^2(x)

    • one year ago
  39. Algebraic! Group Title
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    2a*cos^2(x) - 2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)

    • one year ago
  40. Algebraic! Group Title
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    hmm, I see your point ...

    • one year ago
  41. Algebraic! Group Title
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    I guess you'll be needing something like yp = a*cos^2(x) + b*sin^2(x)

    • one year ago
  42. Algebraic! Group Title
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    let me try it out and see...

    • one year ago
  43. lilMissMindset Group Title
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    sure. :)

    • one year ago
  44. Algebraic! Group Title
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    seems to work... I got a= 2 b=-2

    • one year ago
  45. Algebraic! Group Title
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    so 2cos^2 x -2sin^2 x

    • one year ago
  46. lilMissMindset Group Title
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    answer here in the book is 1/2(1-xsin2x) for the Yp part.

    • one year ago
  47. Algebraic! Group Title
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    I'll assume that's (1/2)*( 1 - x*sin^2 x )

    • one year ago
  48. lilMissMindset Group Title
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    yes.

    • one year ago
  49. Algebraic! Group Title
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    you should check the original, I think you might have written the problem wrong...

    • one year ago
  50. Algebraic! Group Title
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    you left out and x I think...

    • one year ago
  51. lilMissMindset Group Title
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    i don't think so.

    • one year ago
  52. Algebraic! Group Title
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    (1/2)*( sin^2 x - x*sin 2x ) ??

    • one year ago
  53. Algebraic! Group Title
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    I don't see how (1/2)(1-xsin2x) can work... maybe I'm missing something.

    • one year ago
  54. Algebraic! Group Title
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    something's off in your original equation or that soln.

    • one year ago
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