Find the general solution of the given differential equation, and use it to determine how solutions behave as \(t\rightarrow\infty\)
\(y\prime + (1/t)y = 3\cos(2t)\) , \(t>0\)

- austinL

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- austinL

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

- amistre64

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

- austinL

Ok, thanks very much!

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## More answers

- amistre64

i know the homogenous is: c/t

- amistre64

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.

- austinL

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

- amistre64

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

- amistre64

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

- amistre64

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

- austinL

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.

- austinL

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?

- amistre64

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

- austinL

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

- amistre64

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 :)

- amistre64

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

- austinL

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

- austinL

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

- abb0t

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

- abb0t

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).

- austinL

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
\(\mu(t)=t\) ?

- abb0t

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

- abb0t

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

- abb0t

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

- abb0t

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.

- austinL

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

- abb0t

Yes, sir.

- austinL

I see why I am integrating this side :P

- austinL

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

- anonymous

am I too late to this party?

- abb0t

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

- anonymous

Find the general solution of the given differential equation, and use it to determine how solutions behave as tââ
$$y'+\frac1ty=3\cos(2t)$$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:$$\int\frac{d}{dt}[ty]\,dt=\int 3\cos2t\,dt\\ty=3\int\cos2t\,dt=\frac32\sin(2t)+C\\y(t)=\frac3{2t}\sin2t+\frac{C}t$$

- austinL

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

- anonymous

lol very good point

- austinL

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

- austinL

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?

- austinL

+C on end.... oops

- austinL

+C on end.... oops

- abb0t

C'mon gaiz...

- abb0t

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

- anonymous

i prefer jets

- austinL

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?

- abb0t

Ok mr.Fancy pants.

- abb0t

and yes, that is correct.

- austinL

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

- abb0t

Take the limit and find out.

- anonymous

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

- austinL

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

- anonymous

that's because wolfram spit out lim sup/inf

- abb0t

smh @wolfram

- austinL

Zero then, go with my first answer?

- austinL

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?

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