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Wislar
 3 years 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
Wislar
 3 years 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

This Question is Closed

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1o.o first order ODE :o

Wislar
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

Wislar
 3 years ago
Best ResponseYou've already chosen the best response.0I think it has something to do with the solution changing as t>infinity

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1as t approacs infinity the other two expressions approach zero.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1Is it (3/(4t))cos(2t) or (3t/4)cos(2t)?

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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1essentially 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.

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

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1I'm not particularly sure either, but I will try to make sense out of it.

Wislar
 3 years ago
Best ResponseYou've already chosen the best response.0Alright, well thank you!

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1There 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.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.
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