The question in my textbook asks:
If $f$ is a continuous function such that $$\int\limits_0^x{f(t)dt}=xe^{2x}+\int\limits_0^x{e^{-t}f(t)dt}$$ for all $x$, find an explicit formula for $f(x)$.
My working goes as follows:
I decided to analyse the equation as an integration by parts $\left(\int udv=uv-\int vdu\right)$, so
$uv]_0^x=xe^{2x}\\
\therefore \text{a possible substitution is}\\
\quad u=t,\qquad v=e^{2t}\\
\quad du=dt,\quad dv=2e^{2t}dt$
and
$\int\limits_0^xvdu=\int\limits_0^x{e^{-t}f(t)dt}\\
\therefore e^{-t}f(t)=e^{2t}\\
\quad f(t)=e^{3t}$
This looks sound until I try equating $\int\limits_0^xudv=\int\limits_0^x f(t)dt$, whereupon I get $f(t)=2te^{2t}$.
I think I don't quite understand what an explicit formula is.
Best Answer
Notice that if you differentiate the equation:
$$f(x) = e^{2x} + 2xe^{2x} + e^{-x}*f(x)$$
$$f(x)(1 - e^{-x}) = e^{2x} + 2xe^{2x}$$
$$f(x) = \frac{e^{2x} + 2xe^{2x}}{1 - e^{-x}}$$