Question

Find the general solution of the given differential equation, and use it to determine how solutions behave as \(t\rightarrow\infty\)

\(y\prime + (1/t)y = 3\cos(2t)\) , \(t>0\)

Answer 1

@amistre64
Would you mind giving some guidance. I think I am probably making this more difficult than it should be.

Answer 2

maybe in like an hour ill have more time to look at this

Answer 3

Ok, thanks very much!

Answer 4

i know the homogenous is: c/t

Answer 5

y = A(t)~t^(-1)

y' = A'(t)~t^(-1) - A(t) t^(-2) ; let the A'(t) = 0 to account for the homogenous solution

-A(t) t^(-2) = 3cos(2t) would most likely cover the particular ... is a thought.

Answer 6

Ok, the general solution of a problem is y=........ right?
I believe we would have to do the \(\mu(t)\) thing right?

Answer 7

im not adept at the u(t) thing .... never tried to work it like that

Answer 8

the general solution would be: y = yh + yp

in this case yh = C/t

yp can be made with guesses, or by using variation of parameters or a wronskian if memory serves

Answer 9

something like a guess of: yp = A cos(2t) + B sin(2t) is the usual run i believe

Answer 10

In my notes I think I just found something that may be helpful.

\(\dfrac{dy}{dt}+ay=g(t)\)

\(y=e^{-at}\int e^{at}g(t)dt+Ce^{-at}\)

And now that I have it all written out, this particular formula only works if a is a constant.
Darn it.

Answer 11

Ok ok. I think I found an example in my book that is somewhat similar to this problem

\(2y\prime +ty=2\)
Which can be written as,
\(y\prime +\dfrac{t}{2}y=1\)

Then using the \(\mu(t)\) method it says that,
\(\mu(t)=e^{\int p(t)dt}=e^{\int \frac{t}{2}dt}=e^{\frac{t^2}{4}}\)

Could I do something similar here?

Answer 12

im just not that proficient or confident in the u(t) method to validate it

Answer 13

Ok, using words I can understand :P, would you mind walking me through your method and see if it is any better?

Answer 14

in my head .... i something like this:

y = A/t

y/t = A/t^2
+ y' = A'/t - A/t^2
------------------
3 cos(2t) = A'/t

A' = 3t cos(2t)

int A' = 3/4 (2t sin(2t) + cos(2t)) which is apprently off by that errant "t" bythe sin according to the wolf :)

Answer 15

they say 2 sin(2t) + cos(2t)/t

Answer 16

I am sorry, I feel rather dumb right now, but I am having problems wrapping my head around this for some reason.

Answer 17

@Psymon Hate to randomly tag, but others have said that you may be able to help?

Answer 18

This is a first order linear ODE.
Start by obtaining your integrating factor.

Answer 19

Once you get your integrating factor, distribute it across the ODE. You will see that you get product rule on the right hand side (guaranteed. unless you do your math wrong).

Answer 20

Ok, so I have:
\(y\prime +(\dfrac{1}{t})y=3\cos(2t)\)
Our integrating factor multiplied across:
\((\mu(t))y\prime+(\dfrac{1}{t})(\mu(t))y=3(\mu(t))\cos(2t)\)

As I stated earlier,
\(\mu(t)=e^{\int p(t)dt}=e^{\int \frac{1}{t}dt}=e^{\ln|t|}\)
Which I believe means that
\(\mu(t)=t\) ?

Answer 21

Yes, that is correct, \(\mu\)(t)= t
Now, distribute \(\mu\)(t) across your ODE.

Answer 22

You get: \(ty' +y = 3tcos(2t)\)

Answer 23

Notice that your left hand side is nothing more than product rule! So you have: \([ty]' = 3tcos(2t)\)

Answer 24

Integrate both sides with respect to \(t\) to solve for \(y\). The left side you have \(\int [yt]'dt\) = t(y(t)), now i WILL LET you integrate the right hand side. You will be using parts.

Answer 25

I am finding \(\int 3t\cos(2t)dt\) ?

Answer 26

Yes, sir.

Answer 27

I see why I am integrating this side :P

Answer 28

\(=\dfrac{3}{4}(2t\sin(2t)+\cos(2t))+C\)
Does this look correct?

Answer 29

am I too late to this party?

Answer 30

You can take over @oldrin.bataku I thought this was algebra 2 but it's not! So, bye.

Answer 31

Find the general solution of the given differential equation, and use it to determine how solutions behave as t→∞
$$y'+\frac1ty=3\cos(2t)$$Recognize (hopefully by inspection) there is an obvious integrating factor \(\mu=t\) to use here:$$ty'+y=3\cos2t\\(ty)'=3\cos2t$$Integrating both sides we get:$$\int\frac{d}{dt}[ty]\,dt=\int 3\cos2t\,dt\\ty=3\int\cos2t\,dt=\frac32\sin(2t)+C\\y(t)=\frac3{2t}\sin2t+\frac{C}t$$

Answer 32

I believe you made an error in your second line of LaTeX, should it not be \(3t\cos(2t)\)?

Answer 33

lol very good point

Answer 34

And may I ask, where does the "y" go on the left side?

Answer 35

Ok, so right now I am at
\(ty=\dfrac{3}{4}(2t\sin(2t)+\cos(2t))\)

Then do you just divide by t, and it is all good?

Answer 36

+C on end.... oops

Answer 37

+C on end.... oops

Answer 38

C'mon gaiz...

Answer 39

can't we all just be on the same tangent plane for once :-)

Answer 40

i prefer jets

Answer 41

Ok, I think I have the answer.

\(y(t)=\dfrac{3(2t\sin(2t)+\cos(2t))}{4t}+\dfrac{C}{t}\)

Does this look even remotely correct?

Answer 42

Ok mr.Fancy pants.

Answer 43

and yes, that is correct.

Answer 44

WOOOOO!!!!!!!

Then as \(t\rightarrow \infty\), does it approach zero? As there are t's in both denominators?

Answer 45

Take the limit and find out.

Answer 46

@austinL all of $$\sin2t,\cos2t,C$$are bounded while \(t\) is not so yes

Answer 47

When I take the limit, I arrive at \(\dfrac{-3}{2}~to~\dfrac{3}{2}\)

Answer 48

that's because wolfram spit out lim sup/inf

Answer 49

smh @wolfram

Answer 50

Zero then, go with my first answer?

Answer 51

I do believe I will work out another problem, close this and post the question and my solution and see if I am correct. Would you guys mind checking that in a few?