Hey guys,

I've been staring at this question for days and I've got to a point where I don't know where to go from...

The question asks

With the HLDE $\displaystyle y''+4y=f(x), 0\leq x \leq 1$ and the boundary conditions $\displaystyle y(0)=0, y'(1) = 0$, find the two sided green's function.

I chosen the general solution to be $\displaystyle y(x) = Asin(2x)+Bcos(2x)$. As it would satisfy the homogeneous HLDE. I know for the two sided green's function to be constructed, I need two functions $\displaystyle u(x), v(x)$ that satisfy the homogeneous HLDE and satisfy the left and right boundary condition respectively.

Using the left hand side boundary condition,

$\displaystyle y(0)=0, \Rightarrow Asin(0) + Bcos(0) = 0$

Which means $\displaystyle u(x) = sin (2x)$ as $\displaystyle B = 0$ and taking $\displaystyle A = 1$.

Using the right boundary condition and the values found above,

$\displaystyle y'(1)= 0 \Rightarrow 2Acos(2) = 0$

This is where my problem occurs. This shows that I have A & B equal to zero for the BVP, but this can't happen, as I won't be able to construct a green's function. Where have I gone wrong?

Thanks for your time.