​

Correct option is A

​

​$$\text{To derive } K(x, t), \text{ the solution to } y''(x) + 2y(x) = 0 \text{ is written as:} \\y(x) = A\sin(\sqrt{2}x) + B\cos(\sqrt{2}x), \\\text{where } A \text{ and } B \text{ are constants determined by the boundary conditions. For the Green's function:} \\K(x, t) \text{ must behave differently depending on whether } t \leq x \text{ or } t \geq x. \\\text{For } 0 \leq t \leq x, \quad K(x, t) \propto t \sin(\sqrt{2}x), \\\text{For } x \leq t \leq 1, \quad K(x, t) \propto x \sin(\sqrt{2}t). \\\text{Thus, the kernel is written as:} \\K(x, t) = \begin{cases} t \sin(\sqrt{2}x), & \text{if } 0 \leq t \leq x, \\x \sin(\sqrt{2}t), & \text{if } x \leq t \leq 1.\end{cases} \\\text{This ensures that } K(x, t) \text{ is continuous and satisfies the symmetry and boundary conditions.}$$ 

​​​

​