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), t0

Question

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

inkyvoyd · Answer

o.o first order ODE :o

anonymous · Answer

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

anonymous · Answer

I think it has something to do with the solution changing as t->infinity

inkyvoyd · Answer

as t approacs infinity the other two expressions approach zero.

inkyvoyd · Answer

Is it (3/(4t))cos(2t) or (3t/4)cos(2t)?

anonymous · Answer

It is (3/(4t))cos(2t)

inkyvoyd · Answer

$\large \lim_{t->\inf} (3/2)\sin(2t)+\frac{3}{4t}\cos(2t)+(c/t)$
$=(3/2)\sin(2t)+0*\cos(2t)+0$

inkyvoyd · Answer

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.

anonymous · Answer

Thank you! If I just take the limit as t->infinity, then what is the point of adding that t>0?

inkyvoyd · Answer

I'm not particularly sure either, but I will try to make sense out of it.

anonymous · Answer

Alright, well thank you!

inkyvoyd · Answer

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.

inkyvoyd · Answer

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.

anonymous · Answer

thank you!

inkyvoyd · Answer

you're welcome :)