Finding $f(x)$. If $$f(x)=1+x+x^2+\displaystyle\int_{0}^{x}e^k f(x-k) dk$$ then how do we find the function $f(x)$?
Is there a way to solve it, with or without arriving at a differential equation?
This a homework question, the only one i am not able to solve.
 A: Hint: changing variables in the integral $\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k$ by $w=x-k$, $dw=-dk$, makes it easier to differentiate. The substitution gives us:
$$\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k=\int_{0}^{x}e^{x-w}f(w)\,\mathrm{d}w.$$
Differentiating with respect to $x$ on both sides, and applying the general Leibniz rule to the RHS, we ahave:
$$\frac{d}{dx}\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k=\frac{d}{dx}\int_{0}^{x}e^{x-w}f(w)\,\mathrm{d}w\\
=f(x)+\int_{0}^{x}e^{x-w}f(w)\,\mathrm{d}w\\
=f(x)+\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k.$$
Thus, differentiating the integral equation yields an ODE:
$$f'(x)=1+2x+f(x)+\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k$$
$$f'(x)=1+2x+f(x)+\left(f(x)-1-x-x^2\right)$$
$$f'(x)=2f(x)+x-x^2$$
A: Change $x-k$ to $t$ in your integral. Your  equation is now
$$f(x)=1+x+x^2+\exp(x)\int_0^ {x}\exp(-t)f(t)dt$$
Put now $F(x)=\int_0^x \exp(-t)f(t)dt$. You have $F(0)=0$, and your equation is now
$$F^{\prime}(x)=(1+x+x^2)\exp(-x)+F(x)$$
And last hint: Note that $(F^{\prime}-F)\exp(-x)=(F(x)\exp(-x))^{\prime}$
