Differential Equations Initial Value Problem
\(ty\prime +2y=\sin(t)\)
\(y(\frac{\pi}{2})=1,~t>0\)

- austinL

Differential Equations Initial Value Problem
\(ty\prime +2y=\sin(t)\)
\(y(\frac{\pi}{2})=1,~t>0\)

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- terenzreignz

Let's see what we can work out...
\[\Large y ' + \frac2t\cdot y = \frac{\sin(t)}t\]

- terenzreignz

Doesn't seem to work out quite as intended...

- austinL

Ok, so if we were to use the constant of integration it would be \(\mu(t)=2t\)?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- terenzreignz

Oh yeah, it does LOL

- terenzreignz

Okay, so we multiply everything by 2t...
\[\Large 2ty'+ 4y = 2\sin(t)\]

- terenzreignz

You can do it from here?

- austinL

I was just learning this before the weekend and I haven't touched it in 3 days....

- austinL

Do I integrate both sides? What do I do with the 4y? I am confusing myself.

- terenzreignz

Okay... well, you'll notice something about the left side... it's just
\[\LARGE \frac{d}{dt}(2ty)=2\sin(t)\]

- terenzreignz

Using the product rule and implicitly differentiating 2ty with respect to t, you get
2ty' + 4y
^_^

- terenzreignz

And now, just 'multiply both sides' by dt and you'll be on your merry way :D
\[\Large d(2ty) = 2\sin(t)dt\]

- terenzreignz

See where this is going? ^_^
\[\Large \int d(2ty) = \int 2\sin(t)dt\]

- terenzreignz

Finish it !
:D

- austinL

Indeed.

- terenzreignz

So you should get \[\Large 2ty = -2\cos(t) + C\]

- terenzreignz

Now we just need the value of that constant... that's not going to be too hard, right? :D

- austinL

We would need to solve for y right?

- terenzreignz

No... just C :)
You know that
\[\Large y\left(\frac \pi 2\right)= 1\]
So, just have y = 1 and \[\large t = \frac \pi 2\]
and solve for C

- terenzreignz

And then after you find out exactly what value C takes, then you can do the finishing touches and solve for y so that it's a function.
And you'd be done :D

- austinL

\(C=\pi\)

- terenzreignz

Okay :D
So we have
\[\Large 2ty = -2\cos(t) + \pi\]

- austinL

\(y=\dfrac{-\cos(t)}{t}+\dfrac{\pi}{2t}\)
Does this look even remotely correct?

- terenzreignz

Rather, to make things look prettier...
\[\Large 2ty = \pi - 2\cos(t)\]

- terenzreignz

I think this is better :
\[\Large y(t) = \frac{\pi - 2\cos(t)}{2t}\]

- austinL

Okay, but a quick question. On a test, would I have been technically correct?

- terenzreignz

With this:
\[y=\dfrac{-\cos(t)}{t}+\dfrac{\pi}{2t}\]?
Yes. Of course.
I just like it when there's only one fraction bar :D

- austinL

Okay, I just want to make sure! :D

- terenzreignz

hang on a sec..

- austinL

wanted*

- austinL

Thank you very much for your help!!

- terenzreignz

Yeah, just that Wolfie is giving me doubts...

- austinL

My book lists this as it's answer.
\(y=t^{-2}[~(\pi^2/4)-1-t \cos(t) +\sin(t)~]\)

- terenzreignz

Yeah... my bad, wrong integrating factor.

- terenzreignz

This is the correct integrating factor:\[\Large e^{\mu}\]
where
\[\Large \mu = \int \frac2tdt\]

- terenzreignz

This just happens to be \[\Large \mu = 2\int \frac1tdt = 2\ln(t) = \ln (t^2)\]

- austinL

\(\mu(t)=t^2\) I believe.

- terenzreignz

Yup... you and your 2t -.-
LOL
Okay, so THAT's what we multiply to both sides...
\[\Large t^2 y ' + 4ty = t\sin(t)\]

- terenzreignz

---... I think I'm missing something...

- austinL

I think I know.

- terenzreignz

Oh yeah, \[\Large t^2y′+\color{red}2ty=t\sin(t)\]

- austinL

\(t^2y\prime +2ty=2\sin(t)\)

- austinL

Darn it :P

- terenzreignz

So now, we have the left side as
\[\Large \frac{d}{dt}(t^2y)= t\sin(t)\]

- austinL

In all instances, will the middle term just disappear?

- terenzreignz

Yes, that's kind of the point of the integrating factor...

- austinL

Ok, sorry. I must seem so silly right now :P

- terenzreignz

Say, this is general...
\[\Large y' + p'(t)y= q(t)\]

- terenzreignz

Now, the integrating factor would be \[\Large e^{p(t)}\]

- terenzreignz

Multiplying everything through with this factor gives \[\Large e^{p(t)}y' + e^{p(t)}p'(t)y = q(t)e^{p(t)}\]

- terenzreignz

And that thing on the left ALWAYS ends up as:
\[\Large \frac{d}{dt}[e^{p(t)}y]= q(t)e^{p(t)}\]

- terenzreignz

And hopefully, that term on the right (which is a function of t only) is easy to integrate :D

- austinL

Okay, thanks very much! That makes much more sense. My professor wasn't entirely clear about that :)

- terenzreignz

Shall we continue? :)

- austinL

I have \(t^2y=\sin(t)-t\cos(t)+C\)

- austinL

\(C=\dfrac{\pi^2}{4}-1\)

- terenzreignz

And hopefully, you can see where this is going now? ^_^

- austinL

Now for a final answer in a very rough form I have,
\(y=\dfrac{\sin(t)}{t^2}-\dfrac{\cos(t)}{t}+\dfrac{\pi^2}{4t^2}-\dfrac{1}{t^2}\)

- austinL

That can be cleaned up I am sure.

- terenzreignz

But that is the correct answer, right?

- austinL

I believe so.

- terenzreignz

Then... let's get ice cream... to celebrate a job well-done :D

- austinL

Woohoo!!!

- austinL

You deserve many more medals than I can give :P

Looking for something else?

Not the answer you are looking for? Search for more explanations.