I got general solution for PDE I solved
$$f(x, y) = \sum_{n=0}^{\infty} \exp \left(-A \frac{n^2 \pi^2}{4L}y\right)\left[ B_n \cos\left(\frac{2n\pi}{2L}x \right) + C_n\sin\left(\frac{(2n+1)\pi}{2L}x \right)\right] $$
with boundary contition
$$f(x, 0) = N\delta(x-x_0)$$
where $A \in \mathbb{R}_+$, $n \in \mathbb{Z}$, and $B_n$ and $C_n$ are constants related to $n$, $2L$ is length of '$x$' interval where $f(x, y)$ is defined. $\delta(x, x_0)$ is Dirac's Delta
$x \in [-L; L]$
$y \in [0; \infty)$
I'm hoping to designate $B_n$ and $C_n$, but as far as I know I can not get values of two constants from one boundary condition.
What should I do? Can I limit my sum only to odd/even numbers (then either 'sin' term, or 'cos' will survive) without losing generality of my solution?
If so, how should I proceed then?
POSSIBLE SOLUTION Setting $x_0 = 0$ I integrate both sides of $$f(x, 0) = \sum_{n=0}^{\infty} \left[ B_n \cos\left(\frac{2n\pi}{2L}x \right) + C_n\sin\left(\frac{(2n+1)\pi}{2L}x \right)\right]$$ with $$\int dx \cos\left(\frac{2n\pi}{2L}x \right)$$ Using Delta proprety LHS is $$N \int dx \delta(x) \cos\left(\frac{2n\pi}{2L}x \right) = N$$ RHS using orthogonality of trigonometric functions(since $2n+1$ isn't equal to $2m$) will be $$\sum_{n=0}^{\infty} \frac{B_n L}{2 \pi}\delta_{nm} $$ where $\delta_nm$ is Kronecker's delta. Summing with delta leads to $$N = \frac{B_m L}{2 \pi} \rightarrow B_m = N \frac{2\pi}{L}$$
Analogously integrating with $$\int dx \sin\left(\frac{(2m+1)\pi}{2L}x \right)$$ will lead to $$0 = \frac{C_m L}{2 \pi} \rightarrow C_m = 0$$
Then, final soultion will be $$f(x, y) = \sum_{n=0}^{\infty} \exp \left(-A \frac{n^2 \pi^2}{4L}y\right)\left[ N \frac{2\pi}{L} \cos\left(\frac{2n\pi}{2L}x \right)\right] $$