Gotta BVP here that I'd like to run by someone to see if my thinking is correct.

Show that the following boundary value problem:

\(\displaystyle y'' = f(x) (0 \leq x \leq 1)\)

\(\displaystyle y'(0) = 0, y'(1) = 0 \)

does not in general have a solution.

I used the LDE solution to be \(\displaystyle y = Asin(x)+Bcos(x)\).

So, solving for both boundary conditions:

\(\displaystyle y'(0) = Acos(0)-Bsin(0) =0\) Therefore \(\displaystyle A=0\).

\(\displaystyle y'(1) = Acos(1)-Bsin(1) = -Bsin(1) = 0\) Therefore \(\displaystyle B=0\)

This, to me, shows that there is no solution of this form to the BVP. Is this sufficient to answer the question?

The next part asks to find a condition of \(\displaystyle f(x)\) for there to be a solution. If choose the boundary conditions of \(\displaystyle 0\leq x \leq \pi\), and making the right boundary condition to be \(\displaystyle y'(\pi)=0\), I'll get a solution of the form \(\displaystyle y=Bcos(x)\).

Am I close? Or have I got the wrong idea here.

Thanks for your time!