Calculate the value of $f(x)$ $$x^2+\int_{0}^{x}e^{-t}f(x-t)dt=f(x)$$
​
My teacher told me to differentiate this function and then rearrange to integrate.
First, I substituted $x-t=p$;$-dp=dt$.
Therefore,
$$x^2+e^{-x}.\int_{0}^{x}e^{p}f(p)dp=f(x)$$
So, when I applyled Leibniz rule,
In the integrating variable, can I put x instead of p?
That's the point where I got stuck and couldn't figure out what to do.
Can anyone please help me?
 A: Hint:
Before differentiating, transform
$$\int_0^x e^{-t}f(x-t)\,dt=\int_0^x e^{t-x}f(t)\,dt=e^{-x}\int_0^x e^{t}f(t)\,dt$$ and use the Fundamental Theorem of Calculus.
A: Given
$$x^2+\int_{0}^{x}e^{-t}f(x-t)dt=f(x)$$
then by differentiation:
\begin{align}
2 \, x + e^{-x} \, f(0) + \int_{0}^{x} e^{-t} \, f'(x-t) \, dt &= f'(x) \\
2 \, x + f(0) \, e^{-x} + [- e^{-t} \, f(x-t) ]_{0}^{x} - \int_{0}^{x} e^{-t} \, f(x-t) \, dt &= f'(x) \\
2 \, x + f(x) - [ f(x) - x^2 ] &= f'(x) \\
2 \, x + x^2 &= f'(x)
\end{align}
and leads to
$$ f(x) = x^2 + \frac{x^3}{3} + c_{0}.$$
Now, when $x=0$ the original equation yields $f(0) = 0$ and leads to
$$ f(x) = x^2 + \frac{x^3}{3}. $$
As a comparison one can use the convolution of integrals with the Laplace transform. In this view it is determined that:
\begin{align}
\mathcal{L}\{ f(t) \} &= \int_{0}^{\infty} e^{-st} \, f(t) \, dt \\
\frac{2}{s^2} + \mathcal{L}\{e^{-x}\} \, f(s) &= f(s) \\
\frac{2}{s^2} + \frac{f(s)}{s+1} &= f(s) \\
f(s) &= \frac{2(s+1)}{s^4} = \frac{2}{s^3} + \frac{2}{s^4}
\end{align}
and the inverse transform yields
$$ f(x) = x^2 + \frac{x^3}{3}. $$
The two methods end with the same solution.
A: I found out another method;
$$x^2+e^{-x}.\int_{0}^{x}e^{p}f(p)dp=f(x)$$
By applying product rule of differentiation,
$$f(x).g(x)=f'(x)g(x)+g'(x)f(x)$$
We get,
$$2x-e^{-x}.\int_{0}^{x}e^{p}f(p)dp+f(x)=f'(x)$$
Since, $$e^{-x}.\int_{0}^{x}e^{p}f(p)dp=f(x)-x^{2}$$
Therefore,  after substitution we get,
$$2x-(f(x)-x^2)+f(x)=f'(x)$$
$$f'(x)=2x+x^2$$
Therefore,
$$f(x)=x^{2}+\frac{x^{3}}{3}$$
And here, $c=0$ as $f(0)=0$;
