Example 3. Find the Green's function for the boundary value problem

$$
y^{\prime \prime}(x)+k^{2} y=0, \quad y(0)=y(1)=0
$$
Solution. It is easy to verify that the solution $y_{1}(x)=\sin k x$ satisfies the boundary condition $y_{1}(0)=0$, and the solution $y_{2}(x)=\sin k(x-1)$ satisfies the condition $y_{2}(1)=0$, and that they are linearly independent. Let's find the value of the Wronskian determinant for $\sin k x$ and $\sin k(x-1)$ at the point $x=\xi$:

$W(\xi)=\left|\begin{array}{cc}\sin k \xi & \sin k(\xi-1) \\ k \cos k \xi & k \cos k(\xi-1)\end{array}\right|=k[\sin k \xi \cos k(\xi-1)-\sin k(\xi-1) \cos k \xi]=k \sin k$.

Noting that in our example $p(x)=1$, according to (12) we get

$$
G(x, \xi)= \begin{cases}\frac{\sin k(\xi-1) \sin k x}{k \sin k}, & 0 \leqslant x \leqslant \xi \\ \frac{\sin k \xi \cdot \sin k(x-1)}{k \sin k}, & \xi \leqslant x \leqslant 1\end{cases}
$$