# Is there a mistake in the assignment?

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1. Oct 17, 2016

### kostoglotov

1. The problem statement, all variables and given/known data

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

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?

2. Relevant equations

3. The attempt at a solution

2. Oct 17, 2016

### Staff: Mentor

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: Oct 18, 2016
3. Oct 18, 2016

### LCKurtz

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.

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: Oct 18, 2016
4. Oct 18, 2016

### pasmith

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

5. Oct 18, 2016

### LCKurtz

 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:

Last edited: Oct 18, 2016