austinL
  • austinL
Differential Equations Initial Value Problem \(ty\prime +2y=\sin(t)\) \(y(\frac{\pi}{2})=1,~t>0\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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terenzreignz
  • terenzreignz
Let's see what we can work out... \[\Large y ' + \frac2t\cdot y = \frac{\sin(t)}t\]
terenzreignz
  • terenzreignz
Doesn't seem to work out quite as intended...
austinL
  • austinL
Ok, so if we were to use the constant of integration it would be \(\mu(t)=2t\)?

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terenzreignz
  • terenzreignz
Oh yeah, it does LOL
terenzreignz
  • terenzreignz
Okay, so we multiply everything by 2t... \[\Large 2ty'+ 4y = 2\sin(t)\]
terenzreignz
  • terenzreignz
You can do it from here?
austinL
  • austinL
I was just learning this before the weekend and I haven't touched it in 3 days....
austinL
  • austinL
Do I integrate both sides? What do I do with the 4y? I am confusing myself.
terenzreignz
  • terenzreignz
Okay... well, you'll notice something about the left side... it's just \[\LARGE \frac{d}{dt}(2ty)=2\sin(t)\]
terenzreignz
  • terenzreignz
Using the product rule and implicitly differentiating 2ty with respect to t, you get 2ty' + 4y ^_^
terenzreignz
  • 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
  • terenzreignz
See where this is going? ^_^ \[\Large \int d(2ty) = \int 2\sin(t)dt\]
terenzreignz
  • terenzreignz
Finish it ! :D
austinL
  • austinL
Indeed.
terenzreignz
  • terenzreignz
So you should get \[\Large 2ty = -2\cos(t) + C\]
terenzreignz
  • terenzreignz
Now we just need the value of that constant... that's not going to be too hard, right? :D
austinL
  • austinL
We would need to solve for y right?
terenzreignz
  • 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
  • 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
  • austinL
\(C=\pi\)
terenzreignz
  • terenzreignz
Okay :D So we have \[\Large 2ty = -2\cos(t) + \pi\]
austinL
  • austinL
\(y=\dfrac{-\cos(t)}{t}+\dfrac{\pi}{2t}\) Does this look even remotely correct?
terenzreignz
  • terenzreignz
Rather, to make things look prettier... \[\Large 2ty = \pi - 2\cos(t)\]
terenzreignz
  • terenzreignz
I think this is better : \[\Large y(t) = \frac{\pi - 2\cos(t)}{2t}\]
austinL
  • austinL
Okay, but a quick question. On a test, would I have been technically correct?
terenzreignz
  • 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
  • austinL
Okay, I just want to make sure! :D
terenzreignz
  • terenzreignz
hang on a sec..
austinL
  • austinL
wanted*
austinL
  • austinL
Thank you very much for your help!!
terenzreignz
  • terenzreignz
Yeah, just that Wolfie is giving me doubts...
austinL
  • austinL
My book lists this as it's answer. \(y=t^{-2}[~(\pi^2/4)-1-t \cos(t) +\sin(t)~]\)
terenzreignz
  • terenzreignz
Yeah... my bad, wrong integrating factor.
terenzreignz
  • terenzreignz
This is the correct integrating factor:\[\Large e^{\mu}\] where \[\Large \mu = \int \frac2tdt\]
terenzreignz
  • terenzreignz
This just happens to be \[\Large \mu = 2\int \frac1tdt = 2\ln(t) = \ln (t^2)\]
austinL
  • austinL
\(\mu(t)=t^2\) I believe.
terenzreignz
  • 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
  • terenzreignz
---... I think I'm missing something...
austinL
  • austinL
I think I know.
terenzreignz
  • terenzreignz
Oh yeah, \[\Large t^2y′+\color{red}2ty=t\sin(t)\]
austinL
  • austinL
\(t^2y\prime +2ty=2\sin(t)\)
austinL
  • austinL
Darn it :P
terenzreignz
  • terenzreignz
So now, we have the left side as \[\Large \frac{d}{dt}(t^2y)= t\sin(t)\]
austinL
  • austinL
In all instances, will the middle term just disappear?
terenzreignz
  • terenzreignz
Yes, that's kind of the point of the integrating factor...
austinL
  • austinL
Ok, sorry. I must seem so silly right now :P
terenzreignz
  • terenzreignz
Say, this is general... \[\Large y' + p'(t)y= q(t)\]
terenzreignz
  • terenzreignz
Now, the integrating factor would be \[\Large e^{p(t)}\]
terenzreignz
  • 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
  • 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
  • terenzreignz
And hopefully, that term on the right (which is a function of t only) is easy to integrate :D
austinL
  • austinL
Okay, thanks very much! That makes much more sense. My professor wasn't entirely clear about that :)
terenzreignz
  • terenzreignz
Shall we continue? :)
austinL
  • austinL
I have \(t^2y=\sin(t)-t\cos(t)+C\)
austinL
  • austinL
\(C=\dfrac{\pi^2}{4}-1\)
terenzreignz
  • terenzreignz
And hopefully, you can see where this is going now? ^_^
austinL
  • 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
  • austinL
That can be cleaned up I am sure.
terenzreignz
  • terenzreignz
But that is the correct answer, right?
austinL
  • austinL
I believe so.
terenzreignz
  • terenzreignz
Then... let's get ice cream... to celebrate a job well-done :D
austinL
  • austinL
Woohoo!!!
austinL
  • austinL
You deserve many more medals than I can give :P

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