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Homework Help Overview

The discussion revolves around a differential equation with initial conditions, specifically focusing on the constants derived from the general solution y(t) = c_1sin(3t) + c_2cos(3t). The original poster expresses difficulty in determining the constants due to conflicting results from the initial conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the initial conditions and question whether they lead to a valid solution. Some suggest that the conditions may not be appropriate for yielding non-trivial solutions.

Discussion Status

The discussion is ongoing, with some participants providing insights into Sturm-Liouville theory and its relevance to the problem. There is recognition that the initial conditions may lead to a trivial solution, but no consensus has been reached regarding the intent of the problem or the validity of the conditions.

Contextual Notes

Participants note that the initial conditions may not correspond to the expected eigenvalues for the differential equation, raising questions about the problem's setup. There is also mention of potential corrections to earlier statements regarding the eigenvalue associated with the problem.

kostoglotov
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Homework Statement



I'm actually a tutor, and a student of mine at uni has the following differential equation with initial conditions to solve

ptuymQv.gif


imgur link: http://i.imgur.com/ptuymQv.gif

From y(t) = c_1sin(3t) + c_2cos(3t), it is not possible to solve for the constants using the given initial conditions. I always wind up with -c_2 = -3c_1 and c_2 = -3c_1.

Am I missing something or are those init conditions NP?

Homework Equations

The Attempt at a Solution

 
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kostoglotov said:

Homework Statement



I'm actually a tutor, and a student of mine at uni has the following differential equation with initial conditions to solve

ptuymQv.gif


imgur link: http://i.imgur.com/ptuymQv.gif

From y(t) = c_1sin(3t) + c_2cos(3t), it is not possible to solve for the constants using the given initial conditions. I always wind up with -c_2 = -3c_1 and c_2 = -3c_1.

Am I missing something or are those init conditions NP?

Homework Equations

The Attempt at a Solution

If you continue with your work, you get ##c_1 = c_2 = 0##. That means that the solution to the DE is ##y(t) = 0\sin(3t) + 0\cos(3t) \equiv 0##.
That might not have been what the author of the problem intended, but it's a valid solution.
 
Last edited:
This response may well be more advanced than what the OP needs, but I'm giving it for others that might read this thread.

Those boundary conditions look like Sturm-Liouville type B.C.'s, and you wouldn't normally expect a non-zero solution for any old value of the constant, which is ##9## in this problem. For example, consider the eigenvalue problem$$
y''+\mu^2 y = 0$$ $$y(0)+y'(0) = 0,~~y(\pi)-y'(\pi)=0$$Sturm-Liouville theory tells us there should be infinitely many eigenvalues ##\lambda_n=\mu_n^2##, with corresponding non-zero eigenfunctions (solutions). Is one of the eigenvalues ##\lambda = 9##? The answer is "no". Without boring you with the details, if you work out the details, the ##\mu_n## are solutions of the equation$$\tan(\pi \mu) = \frac{2\mu}{1-\mu^2}$$They can be found numerically and are where the functions ##\tan(\pi \mu)## and ##\frac{2\mu}{1-\mu^2}## in the picture below cross.
eigenvalues.jpg


Unfortunately, ##9## isn't one of the eigenvalues, which is why there is only the trivial solution.
[Edit, added]: Note ##\lambda = 9## would correspond to ##\mu = 3##.
 
Last edited:
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LCKurtz said:
View attachment 107655

Unfortunately, ##9## isn't one of the eigenvalues, which is why there is only the trivial solution.

Unfortunate that the graph doesn't clearly demonstrate this by having the horizontal axis include \mu = 9.
 
pasmith said:
Unfortunate that the graph doesn't clearly demonstrate this by having the horizontal axis include \mu = 9.

[Edit] This may need correcting, stay tuned...
OK, the below is corrected.
@pasmith: I was apparently asleep when I first replied. For ##\lambda = 9## we need ##\mu## near ##3##, which is illustrated in my previous post. The ##\mu## value nearest ##3## is 2.780188423. This leads to a solution of y = sin(2.780188423*x)-2.780188423*cos(2.780188423*x). Here's a graph:

solution.jpg
 
Last edited:

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