austinL
  • austinL
Differential Equation Answer Check... help likely needed as well. Find the general solution of the given differential equation, and use it to determine how solutions behave as \(t\rightarrow \infty\) \(2y\prime+y=3t^2\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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austinL
  • austinL
The answer that I arrived at is: \(y(t)=e^{\frac{-1}{2}t}(3e^{\frac{t}{2}}(t^2-4t+8))+Ce^{\frac{-1}{2}t}\) However, I feel like this should be able to be reduced somehow.
abb0t
  • abb0t
Yes, you could still simplyify it a bit more to make it look neater. Lol
austinL
  • austinL
How on earth.... I feel like such a schmuck for asking so many questions that seem so simple. How would one set about simplifying this?

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abb0t
  • abb0t
distribute the 3e\(^{\frac{t}{2}}\)...then the e\(^{-\frac{t}{2}}\)
austinL
  • austinL
Wait... Would the \(e^{\frac{-t}{2}}\)cancel out the \(e^{\frac{t}{2}}\), and then Leave just the 3 and that would distribute?
abb0t
  • abb0t
You \(\sf \color{red}{don't}\) NEED to simplyify tho, this isn't an algebra course, BUT just easier to read if you're doing this on an exam.
abb0t
  • abb0t
I was just saying you could of simplified it a bit for me! lol.
austinL
  • austinL
\(y(t)=3t^2-12t+24+Ce^{\frac{-t}{2}}\)
abb0t
  • abb0t
yes.
abb0t
  • abb0t
You shouldn't be struggling on simplification tho at this level of math.
Loser66
  • Loser66
something wrong above , please check. where is your int?
austinL
  • austinL
I am having a small series of brain farts ok? And, what is wrong??
abb0t
  • abb0t
I think he did his work on his paper. There's no need to show integral here. The integral is one of the basic integrals and is able to be integrated.
abb0t
  • abb0t
because some answers are not integrable and must be left as an integral at this level of mathematics.
austinL
  • austinL
I know that, I have some of those in my notes. However, this one is integrable, and as such isn't a necessary part of the answer right?
Loser66
  • Loser66
to me, \[y = e^{-\frac{1}{2}t}\int 3t^2e^{\frac{1}{2}t} + C e^{\frac{1}{2}t}\]
Loser66
  • Loser66
and for the int. you have to use tabular to solve . I mean integral by part twice
austinL
  • austinL
Ok.
Loser66
  • Loser66
cannot stay, my "trash " computer doesn't work well
abb0t
  • abb0t
Which ever method he used, he arrived at the correct solution. Plus, I am not his grader so I have no idea whether he was doing it correctly to give him "full credit". So I have no say in his methodology.
austinL
  • austinL
Ok, well... I have to continue working out problems. Should I keep posting them here? Or shall I start new questions each time? I don't quite feel confident yet.
abb0t
  • abb0t
he could of used laplace, or series to solve this, or used the integrating factor method.
Loser66
  • Loser66
@abb0t . I am new in this field. I have to be careful
abb0t
  • abb0t
It's ok, Loser. If you have questions, post them up. There are a lot of users on here who can help in ODE's. I am probably not that person. But i can recommend some who know this math course.
austinL
  • austinL
And to clarify for you @abb0t we are coving integrating factors right now, so I am to assume that is the method he would expect us to use.
austinL
  • austinL
And to clarify for you @abb0t we are coving integrating factors right now, so I am to assume that is the method he would expect us to use.
abb0t
  • abb0t
Ok.
austinL
  • austinL
My next problem is an Initial Value Problem. \(y\prime -y=2t~e^{2t}\) \(y(0)=1\) \(y(t)=\dfrac{2(3t-1)e^{2t}}{9}+Ce^t\) Then you would plug in zero for t, and then solve for C right?
austinL
  • austinL
Anyone have any input?
Loser66
  • Loser66
I don't check your stuff, just answer your question. Yes,
austinL
  • austinL
Then, do you plug what you have back in for C, and then that is your final answer?
Loser66
  • Loser66
Again, Yes
Loser66
  • Loser66
Are you solving physics part? find velocity? distance? or spring mass?
austinL
  • austinL
Nope, it is just a plain jane initial value problem.
Loser66
  • Loser66
ok, good luck.
austinL
  • austinL
Thanks, I think I have it!

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