Psymon
  • Psymon
Small issue with the last part of this diff eq problem: The question is a piece-wise and wants me to determine the value of c such that the function is continuous. I'll draw the piecewise
Mathematics
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SOLVED
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chestercat
  • chestercat
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Psymon
  • Psymon
\[\frac{ dy }{ dx }+2y = f(x); y(0) = 0 \] |dw:1378869728679:dw|
Psymon
  • Psymon
Doing the diff eq is no problem, I just dont understand how the book came up with its value for c that made it defined.
Psymon
  • Psymon
*continuous

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anonymous
  • anonymous
Where's the c ?
Psymon
  • Psymon
Well, the constant after you integrate. After integrating the second portion of the piecewise Im suppose to find c such that I can piece the two solutions together and have it be continuous from [0,infinity)
anonymous
  • anonymous
Did you treat this as a linear ODE or did you apply the Laplace transform?
Psymon
  • Psymon
Linear.
anonymous
  • anonymous
Just a sec, I'll see if working it out with Laplace will shed some light on the situation...
Psymon
  • Psymon
Alrighty, thanks.
anonymous
  • anonymous
At any point, do you get something like this \[y=\frac{1}{2}\left(1-e^{6-2x}\right)+Ce^{-2x}~~?\]
Psymon
  • Psymon
Yeah, I just dont have the 6 I guess. Because I did it as if it were linear. After solving for the second part of the piecewise, i get y = Ce^(-2x) as you did. I just didnt understand how the book got its answer for C as to make it continuous after that. In fact, the first term of your answer looks just like the answerfor the first portion of the piecewise that I got, just no 6 in the exponent.
anonymous
  • anonymous
Ok, so I suspect you have written \[e^{2x}y=\int_0^xe^{2x}~f(x)~dx+C\] Well, \(f(x)=0\) for \(x>3\), so you're left with \[e^{2x}y=\int_0^3e^{2x}~dx+C\\ y=e^{-2x}\left[\frac{1}{2}e^{2x}\right]_0^3+Ce^{-2x}\]
anonymous
  • anonymous
The bracketed quantity becomes \(e^{2x}-e^6\), and multiplied by \(e^{-2x}\) is \(1-e^{6-2x}\).
anonymous
  • anonymous
But now I'm wondering what's meant by "such that the function is continuous"... This result is continuous for any value of C.
Psymon
  • Psymon
Well, I think you just hit on it, it's just the book, as you can imagine some texts do, kind of jumped to conclusions. There definitely was no mention of actually applying limits to the integration. I just integrated as if there were no limits. Ended up with \[\frac{ 1 }{ 2 }(1-e ^{-2x})\]For the first portion and \[Ce ^{-2x} \]For the second. To me it seems like it'd be continuous anyway, but then the book comes up with an answer for C that it looks like you got normally: \[\frac{ 1 }{ 2 }(e ^{6}-1)e ^{-2x} \] So thats all you have to do is actually use 0 to 3 as limits? Considering there isnt anything like that in the one example, I just thought there was some sort of reasoning I wasnt understanding.
anonymous
  • anonymous
Maybe this will explain it: http://www.jirka.org/diffyqs/htmlver/diffyqsse3.html#x7-100001.1 By the way, what is the entire answer that the book provides?
Psymon
  • Psymon
Well, let me know if you need more info than this, but this was the reasoning the book used in the one example given, followed by the answer the book gives. So the book example was: \[\frac{ dy }{ dx }+y = f(x); y(0) = 0\] where f(x) = 1 for 0<= x <= 1 and f(x) = 0 for x >1 So it integrates both portions. It comes up with \[c _{2}e ^{-x}\]and then applies this reasoning: "By appealing to the definition of continuity at a point, it is possible to determine c2 so that the foregoing function is continuous at x = 1. The requirement that lim x->1+, y(x) = y(1) implies that c2^(-1) = 1-e^-1 or c2 = e-1" Now as for the book answer to my question it gives: |dw:1378872487708:dw| Because integrating it normally gave me y=ce^-(2x), I had no idea how it jumped to the conclusion it did of the c being (1/2)(e^6-1)
anonymous
  • anonymous
Hmm, using the method described in the link gives me the same answer as your textbook's example. But it looks like I may have complicated things by introducing that stuff... Given \(f\), you can split up the initial ODE into \[y'+2y=1~~~~~~~~~~~~y'+2y=0\] then solve them individually as you normally would, keeping in mind that the left equation holds for \(0\le x\le3\) and the right for \(x>3\). The left equation yields \(y=\dfrac{1}{2}\left(1-e^{-2x}\right)\), and the right yields \( y=Ce^{-2x}\). Judging by the explanation for the example, you want \(\displaystyle\lim_{x\to3^+}=y(3)\), where \(y(3)=\dfrac{1}{2}\left(1-e^{-6}\right)\), according to the first equation's solution. So you want the following to hold: \[\frac{1}{2}\left(1-e^{-6}\right)=Ce^{-6}~~~~~~~~~~~~\text{(direct substitution on the limit)}\\ \frac{1}{2}\left(e^6-1\right)=C\] Hope that all makes sense. At the very least it agrees with the textbook's solution.
Psymon
  • Psymon
No biggie, I just wasnt catching on to where it exactly it was coming up with certain things. Because everything was in terms of 1's, I guess I didnt see the actual plugging in of a limit, just assume it was some definition thing. But yeah, it makes perfect sense, thanks a lot ^_^
anonymous
  • anonymous
You're welcome! Glad I could help
Psymon
  • Psymon
You always do xD

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