I have the following equation, which I need to solve:
$\left(\frac{\mathrm{d}^2}{\mathrm{d} x^2}-a^2\right)A(x)=c\ \delta(x-b)$.
The solution should be in form $A(x)=C_1 e^{-x a}+C_2 e^{x a}+f(x)$,
but I am not sure with the non-homogeneous part $f(x)$. Mathematica offers the solution with Heaviside theta function
$f(x)=\left[c\ e^{-a b - a x} (-e^{2 a b} + e^{2 a x}) \Theta(-b + x)\right]/(2 a)$,
however in literature I found the solution in form
$f(x)=\left[c\ e^{-a \lvert x-b \rvert}\right]/(2 a)$.
Are these two forms equivalent?
The two forms are equivalent, but with different coefficients. The boundary conditions have to be taken into account in the relationships between the respective coefficients.
$$A''(x)-a^2A(x)=c\delta(x-b)$$ Let : $A(x)=e^{ax}Y(x) \quad\to\quad Y''(x)+2aY'(x)=c \delta(x-b)e^{-ax}$
$Y'(x)=e^{-2ax}y(x) \quad\to\quad y'(x)=c\delta(x-b)e^{ax}$
$$y(x)=\int c\delta(x-b)e^{ax}dx = c\:e^{ab}\theta(x-b)+C_0$$ $\theta$ is the Heaviside step function.
$Y'(x)=e^{-2ax}y(x) = c\:e^{ab}e^{-2ax}\theta(x-b)+Ce^{-2ax}$
$Y(x)= ce^{ab}\int e^{-2ax}\theta(x-b)dx+C_1e^{-2ax}+C_2$
$\int e^{-2ax}\theta(x-b)dx= \frac{1}{2a}\left(e^{-2ab}-e^{-2ax} \right)\theta(x-b)$
$$Y(x)= ce^{ab} \frac{1}{2a}\left(e^{-2ab}-e^{-2ax} \right)\theta(x-b)+C_1e^{-2ax}+C_2$$
$A(x)=e^{ax}Y(x)=ce^{ab} \frac{1}{2a}e^{ax}\left(e^{-2ab}-e^{-2ax} \right)\theta(x-b)+C_1e^{-ax}+C_2e^{ax}$
$$A(x)=c\frac{1}{2a}\left(e^{a(x-b)}-e^{-a(x-b)} \right)\theta(x-b)+C_1e^{-ax}+C_2e^{ax}$$
This result is the same as the Mathematica result.
This is equivalent to : $\quad A(x)=\begin{cases} C_1e^{-ax}+C_2e^{ax} \quad\text{if } x<b\\ C_3e^{-ax}+C_4e^{ax} \quad\text{if } x>b\end{cases}$
because both are function of $e^{ax}$ and $e^{-ax}$ but of course with different constant coefficients depending on the boundary conditions.
Comparison to the expected result : $$\quad A(x)=c\frac{1}{2a}e^{-a|x-b|} +c_1e^{-ax}+c_2e^{ax} $$
$$\quad A(x)= \begin{cases} C'_1e^{-ax}+C'_2 e^{ax} \quad\text{if } x<b\\ C'_3e^{-ax}+C'_4 e^{ax} \quad\text{if } x<b \end{cases}$$ because both are function of $e^{ax}$ and $e^{-ax}$ but with different coefficients depending on the boundary conditions.
Thus, the two forms of results (one from mathematica, the other from the literature) are equivalent, insofar the coefficients are related one to another according to the boundary conditions.