
#1
Mar1412, 04:45 AM

P: 884

1. The problem statement, all variables and given/known data
Find a oneparameter family of solutions to the differential equation: [itex]\frac{dy}{dt} = \frac{t \ cos 2t}{y}[/itex] Are there any solutions to the differential equation that are missing from the set of solutions you found? Explain. 3. The attempt at a solution I used the separation of variables to find the solution as follows: [itex]\int ydy = \int t \ cos (2t) dt[/itex] [itex]\frac{y^2}{2} = \frac{1}{4} (2t \ sin (2t)+ cos(2t)) + c[/itex] [itex]\therefore y = \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}[/itex] And this is defined as long as there is no negative under the square root. So, how do I know whether there are any missing solutions or not? And how do I identify them? 



#2
Mar1412, 01:37 PM

P: 312

There could be a negative sign in front of the square root, with an independent constant c(or k).




#3
Mar1512, 03:49 AM

P: 884





#4
Mar1512, 05:19 AM

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P: 26,167

DE Missing Solutions
hi roam!
if y is a solution, then obviously y is also 



#5
Mar1512, 03:39 PM

P: 884

[itex]\therefore y = \pm \sqrt{t \ sin (2t)+ \frac{1}{2} \ cos (2t) + k}[/itex] .......(1) But this doesn't answer the second part of the question: Are there any solutions to the differential equation that are missing from the set of solutions you found? Are there any solutions to the DE that are missing from the set (1)? 



#6
Mar1512, 04:11 PM

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P: 4,670

RGV 



#7
Mar1512, 04:13 PM

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P: 26,167

(the only dangerous thing we did was to multiply both sides by y, and "y = 0 for all t" obviously isn't a solution in this case) 



#8
Mar1512, 06:50 PM

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#9
Mar1512, 06:57 PM

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P: 26,167

so we check, we find it isn't, and we relax again 



#10
Mar1612, 09:28 PM

P: 884

Hi tiny tim,
I have two questions: 1. If using the "±" symbol is wrong, then what is the correct notation for representing the solution to this DE? 2. So, there are absolutely no missing solutions, and my answer is the general solution and we can solve all initialvalue problems with solutions of this form? (Personally I can't think of any counter examples, but I'm not sure). 



#11
Mar1712, 01:22 AM

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P: 4,670

Anyway, if you look at the IVP and have y(0) > 0, then you must choose the solution y1 (with the + sign); if y(0) < 0 you must choose y2 (with the  sign). Then the issue is: are there any other solutions to your IVP? You can use theorems about that. Google "uniqueness theorems for differential equations". RGV 



#12
Mar1712, 04:13 AM

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hi roam!
if your professor doesn't like it, don't do it if your professor doesn't mind it, do do it it's a convenient shorthand … when we write "the general solution is x = Acoskt + C", we don't feel the need to add "for any constant C" when we write "the general solution is x = b ±√(b^{2}  4c), we don't feel the need to add "for either value of ±" 



#13
Mar1712, 12:26 PM

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P: 4,670

RGV 



#14
Mar1712, 08:16 PM

P: 884

So [itex]\frac{\partial f}{\partial y} =  \frac{t \ cos \ 2t}{y^2}[/itex] When y=0, the partial derivative fails to exist, this means that the uniqueness theorem doesn't tell us anything about solutions of an IVP that have the form y(t0)=0. For example at the point y(0)=0, neither f(t, y) or ∂f/∂y exist. But how can we use this to prove that y(0)=0 is another solution to the IVP? I'm confused because the satisfaction of the theorem simply guarantees a uniques solution, but failure of the theorem doesn't prove multiple solutions (inconclusive). So how can we use this theorem? 



#15
Mar1712, 09:02 PM

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So, if y(0) is not zero you have a unique solution; in fact, you can see from the DE itself that y(t) is strictly increasing if y(0) > 0 and is strictly decreasing if y(0) < 0, so y(t) never reaches zero. The case y(0) = 0 is pathological, and I don't think you can say the DE even means anything right at the point t = 0.
RGV 



#16
Mar1812, 12:37 AM

P: 884

Right, so how else can I find the missing solutions?




#17
Mar1812, 01:57 AM

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P: 4,670

RGV 



#18
Mar1812, 02:24 AM

P: 884




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