# Two sided green's function

#### Silverflow

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...

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?