## Wislar Group Title 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

1. inkyvoyd Group Title

o.o first order ODE :o

2. Wislar Group Title

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

3. Wislar Group Title

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

4. inkyvoyd Group Title

as t approacs infinity the other two expressions approach zero.

5. inkyvoyd Group Title

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

6. Wislar Group Title

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

7. inkyvoyd Group Title

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

8. inkyvoyd Group Title

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.

9. Wislar Group Title

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

10. inkyvoyd Group Title

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

11. Wislar Group Title

Alright, well thank you!

12. inkyvoyd Group Title

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.

13. inkyvoyd Group Title

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.

14. Wislar Group Title

thank you!

15. inkyvoyd Group Title

you're welcome :)