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Wislar

Mostly solved, just help on one step! Find the general solution and use it to determine how solutions behave as t->inf: y'+(1/t)y=3cos(2t), t>0

  • one year ago
  • one year ago

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  1. inkyvoyd
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    o.o first order ODE :o

    • one year ago
  2. Wislar
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    I found the solution as: \[y=(3/2)\sin(2t)+(3/4t)\cos(2t)+(c/t)\] which it says is correct, but then it goes on to say that the general solution is: \[y=(3/2)\sin(2t)\] and I'm not sure how they got that

    • one year ago
  3. Wislar
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    I think it has something to do with the solution changing as t->infinity

    • one year ago
  4. inkyvoyd
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    as t approacs infinity the other two expressions approach zero.

    • one year ago
  5. inkyvoyd
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    Is it (3/(4t))cos(2t) or (3t/4)cos(2t)?

    • one year ago
  6. Wislar
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    It is (3/(4t))cos(2t)

    • one year ago
  7. inkyvoyd
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    \(\large \lim_{t->\inf} (3/2)\sin(2t)+\frac{3}{4t}\cos(2t)+(c/t)\) \(=(3/2)\sin(2t)+0*\cos(2t)+0\)

    • one year ago
  8. inkyvoyd
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    essentially as t approaches infinity we still see oscillating behavior in y, but the effects of the cos term and c/t are greatly reduced to the point they are negligible.

    • one year ago
  9. Wislar
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    Thank you! If I just take the limit as t->infinity, then what is the point of adding that t>0?

    • one year ago
  10. inkyvoyd
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    I'm not particularly sure either, but I will try to make sense out of it.

    • one year ago
  11. Wislar
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    Alright, well thank you!

    • one year ago
  12. inkyvoyd
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    There may be a case where you take the natural logarithm of an expression, and in that case t must be greater than 0. I think that is why they have that restriction.

    • one year ago
  13. inkyvoyd
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    I mean while doing the computation, you must take the natural logarithm with an expression with t or a multiple of t, and if t<=0 the operation is undefined.

    • one year ago
  14. Wislar
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    thank you!

    • one year ago
  15. inkyvoyd
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    you're welcome :)

    • one year ago
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