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lilMissMindset

  • 3 years ago

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

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  1. shubhamsrg
    • 3 years ago
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    y= a cos2x ?

  2. shubhamsrg
    • 3 years ago
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    guess that substn should work..

  3. lilMissMindset
    • 3 years ago
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    aaa. i think i got it. wait, i'll try.

  4. shubhamsrg
    • 3 years ago
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    ohh wait,,i didnt see a "y" also in there. !!

  5. Algebraic!
    • 3 years ago
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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 ?

  6. Algebraic!
    • 3 years ago
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    heh.

  7. shubhamsrg
    • 3 years ago
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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..

  8. Algebraic!
    • 3 years ago
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    \[\frac{ d^2y }{ dx^2 } +4y =4\sin^2x\] rather.

  9. lilMissMindset
    • 3 years ago
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    yes, tat's it

  10. lilMissMindset
    • 3 years ago
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    that*

  11. Algebraic!
    • 3 years ago
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    yes he's correct I believe y =a*cos^2 x

  12. Algebraic!
    • 3 years ago
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    did you get it to work?

  13. Algebraic!
    • 3 years ago
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    that's the particular soln. anyway...

  14. lilMissMindset
    • 3 years ago
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    isn't it like, Acosx+Bsinx+Cxcosx+Dxsinx

  15. Algebraic!
    • 3 years ago
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    why?

  16. lilMissMindset
    • 3 years ago
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    since sinx is twice? applying case 4 of undetermined coefficients

  17. lilMissMindset
    • 3 years ago
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    where it has some repeated imaginary root

  18. Algebraic!
    • 3 years ago
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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)

  19. Algebraic!
    • 3 years ago
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    this is for the particular solution...

  20. Algebraic!
    • 3 years ago
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    finding the homogeneous solution shouldn't be a problem...

  21. Algebraic!
    • 3 years ago
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    \[\lambda ^2 +4 =0\]

  22. Algebraic!
    • 3 years ago
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    lambda = 2i

  23. Algebraic!
    • 3 years ago
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    ok?

  24. lilMissMindset
    • 3 years ago
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    i'll work on it. please guide me still, in case i get stuck.

  25. Algebraic!
    • 3 years ago
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    hold on... can you tell me what part is giving you problems?

  26. lilMissMindset
    • 3 years ago
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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.

  27. Algebraic!
    • 3 years ago
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    ok:)

  28. lilMissMindset
    • 3 years ago
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    how about if the given is cos^x, then Yp=asin^2 x?

  29. Algebraic!
    • 3 years ago
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    if the forcing term is cos^2 x you mean?

  30. lilMissMindset
    • 3 years ago
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    yes.

  31. lilMissMindset
    • 3 years ago
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    i forgot 2, lol.

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

  33. Algebraic!
    • 3 years ago
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    yes. it would. just checked it.

  34. lilMissMindset
    • 3 years ago
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    thanks!

  35. Algebraic!
    • 3 years ago
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    sure:)

  36. lilMissMindset
    • 3 years ago
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    darn, i can't do it. show me a brief solution please. so i can learn how.

  37. lilMissMindset
    • 3 years ago
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    just for the Yp part. swear.

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

  39. Algebraic!
    • 3 years ago
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    2a*cos^2(x) - 2a*sin^2(x) +4a cos^2(x) = 4sin^2(x)

  40. Algebraic!
    • 3 years ago
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    hmm, I see your point ...

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

  42. Algebraic!
    • 3 years ago
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    let me try it out and see...

  43. lilMissMindset
    • 3 years ago
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    sure. :)

  44. Algebraic!
    • 3 years ago
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    seems to work... I got a= 2 b=-2

  45. Algebraic!
    • 3 years ago
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    so 2cos^2 x -2sin^2 x

  46. lilMissMindset
    • 3 years ago
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    answer here in the book is 1/2(1-xsin2x) for the Yp part.

  47. Algebraic!
    • 3 years ago
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    I'll assume that's (1/2)*( 1 - x*sin^2 x )

  48. lilMissMindset
    • 3 years ago
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    yes.

  49. Algebraic!
    • 3 years ago
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    you should check the original, I think you might have written the problem wrong...

  50. Algebraic!
    • 3 years ago
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    you left out and x I think...

  51. lilMissMindset
    • 3 years ago
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    i don't think so.

  52. Algebraic!
    • 3 years ago
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    (1/2)*( sin^2 x - x*sin 2x ) ??

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

  54. Algebraic!
    • 3 years ago
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    something's off in your original equation or that soln.

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