Solve the equation involving integral

Question

Let $$f(x) = e^{-x} + 2\int_0^x e^{-3t}f(x-t)dt,$$ solve $f(x)$.

I tried to apply $u = x - t$ but it didn't help . Also using Leibniz integral rule leads to $$f'(x) = -e^{-x} + 2(e^{-3x} + \int_0^x e^{-3t}f'(x-t)dt)$$ and I'm stuck here .

Answer 1

After applying $u=x-t$ one obtains $$f(x)=e^{-x}+2e^{-3x}\int_0^xe^{3u}f(u)du.$$ Differentiating this equation gives $$f'(x)=-e^{-x}-6e^{-3x}\int_0^xe^{3u}f(u)du+2e^{-3x}\cdot e^{3x}f(x)du.$$ Using the first equation, this can be written as $$f'(x)=-e^{-x}-3(f(x)-e^{-x})+2f(x)$$ which leads to linear differential equation $$f'(x)+f(x)=2e^{-x}$$ whose solutions are $f(x)=(2x+C)e^{-x}$, $C\in\mathbb{R}$.

Inserting expression for $f(x)$ in the first equation and solving for $C$ gives only one possible solution which is $f(x)=(2x+1)e^{-x}$.