I am trying to understand Dirac delta function. Here is a DE to solve: $f''(x) - 3f'(x) +2f(x)=k\delta(x-a)$ with intial conditions $ f(0)=f'(0)=1$ where $ k \in \mathbb{R}$ is constant.
I know this is dead easy to solve with Laplace transform, but how can I do it directly?
Just need a hint, a starting point...
You know the eigenvalues are 1 and 2, thus consider $g(x)=e^{-x}f(x)$ with $$ g'-g=e^{-x}(f'(x)-2f(x)),\\---\\ g''-g'=e^{-x}(f''(x)-3f'(x)+2f(x))=e^{-x}·kδ(x−a)=e^{-a}·kδ(x−a) $$ This can be integrated once, and using the Heaviside function as the integral of delta, $$ g'-g=ke^{-a}·H(x-a)+C $$ Substituting again $h(x)=e^{-x}g(x)$ results in $$ h'(x)=e^{-x}(g'-g)=ke^{-a-x}·H(x-a)+C·e^{-x} $$ with integral $$ h(x)=k(e^{-2a}-e^{-a-x})H(x-a)-C·e^{-x}+D \\ \implies \\ f(x)=e^{2x}·h(x)=k(e^{2(x-a)}-e^{x-a})H(x-a)-C·e^x+D·e^{2x} $$
By playing around with the constant, this can be changed into $$ f(x)=\frac k2·\left|e^{2(x-a)}-e^{x-a}\right|+c·e^x+de^{2x} $$