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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
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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inkyvoydBest ResponseYou've already chosen the best response.1
o.o first order ODE :o
 one year ago

WislarBest ResponseYou've already chosen the best response.0
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

WislarBest ResponseYou've already chosen the best response.0
I think it has something to do with the solution changing as t>infinity
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
as t approacs infinity the other two expressions approach zero.
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
Is it (3/(4t))cos(2t) or (3t/4)cos(2t)?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
\(\large \lim_{t>\inf} (3/2)\sin(2t)+\frac{3}{4t}\cos(2t)+(c/t)\) \(=(3/2)\sin(2t)+0*\cos(2t)+0\)
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
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

WislarBest ResponseYou've already chosen the best response.0
Thank you! If I just take the limit as t>infinity, then what is the point of adding that t>0?
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
I'm not particularly sure either, but I will try to make sense out of it.
 one year ago

WislarBest ResponseYou've already chosen the best response.0
Alright, well thank you!
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.1
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

inkyvoydBest ResponseYou've already chosen the best response.1
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
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