Is there a mistake in the assignment?

In summary, the student's problem has initial conditions that do not allow for a solution and the tutor suggests modifying the conditions to make a solution possible.
  • #1
kostoglotov
234
6

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

Am I missing something or are those init conditions NP?

Homework Equations

The Attempt at a Solution

 
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  • #2
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 [itex]y(t) = c_1sin(3t) + c_2cos(3t)[/itex], it is not possible to solve for the constants using the given initial conditions. I always wind up with [itex]-c_2 = -3c_1[/itex] and [itex]c_2 = -3c_1[/itex].

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:
  • #3
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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Likes pasmith
  • #4
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 [itex]\mu = 9[/itex].
 
  • #5
pasmith said:
Unfortunate that the graph doesn't clearly demonstrate this by having the horizontal axis include [itex]\mu = 9[/itex].

[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
 
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