I'm relatively new to the concept of the Dirac Delta function have come across a problem in dealing with ODE with delta
Solve the ODE:
$$A''(y) - λ^2 A(y) = δ(y - ξ)$$
Subject to B.C (Hint: Use Hyperbolic functions) $$A(0) = A(b) = 0,$$
For the solution I have assumed $δ(ξ)$ is undefined, So $$A(y) = \begin{cases} Asinh(λy)+BCosh(λy), & y<ξ &(1) \\ Csinh(λy)+DCosh(λy), & y>ξ&(2) \end{cases}$$
Using $A(0)=0, \ A(y)=Asinh(λy)$
Using $A(b)=0, \ A(y)=Csinh(λ(b - y))$
Such that; $$A(y) = \begin{cases} Asinh(λy), & y<ξ &(1) \\ Csinh(λ(b - y), & y>ξ&(2) \end{cases}$$
What can I do to further simplify my solution, Can I get the constants A , C?
I'm going to assume that $0 < \xi < b$. Let $\epsilon>0$, then take your differential equation and integrate it like so:
$$\int_{\xi-\epsilon}^{\xi+\epsilon} A''(y) - \lambda^2 A(y) dy = \int_{\xi-\epsilon}^{\xi+\epsilon} \delta(y-\xi)dy$$
$$\implies A'\left(\xi+\epsilon\right) - A'\left(\xi-\epsilon\right) -\lambda^2\int_{\xi-\epsilon}^{\xi+\epsilon} A(y)dy = 1$$
Now take the limit as $\epsilon\to 0^+$. By assumption, we want $A(y)$ to be continuous, so the integral term goes to $0$. What we are left with is the following:
$$A'\left(\xi^+\right)-A'\left(\xi^-\right) = 1$$
This is your new condition to solve for the remaining coefficients. The first derivative is not continuous at $y=\xi$, but it has a jump discontinuity that is exactly $1$ in size.