Finding $f(x)$.

Question

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.

The only two ways I know to solve this are to use Laplace Transforms and to convert that integral equation to a differential equation. — JimmyK4542, Jul 26 '14 at 07:00
See here and follow the links you will find different techniques for solving such integral equation. — Mhenni Benghorbal, Jul 26 '14 at 07:01
Take care ! I have the feeling that I made a mistake somewhere and I do noit find it now. I prefer to advise you immediately. — Claude Leibovici, Jul 26 '14 at 07:25

score 4 · Accepted Answer · answered Jul 26 '14 at 07:16

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:

$$f'(x)=1+2x+f(x)+\int_{0}^{x}e^{k}f(x-k)\,\mathrm{d}k$$

$$f'(x)=1+2x+f(x)+\left(f(x)-1-x-x^2\right)$$

$$f'(x)=2f(x)+x-x^2$$

score 0 · Answer 2 · answered Jul 26 '14 at 07:14

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 $(F^{\prime}-F)\exp(-x)=(F(x)\exp(-x))^{\prime}$

Finding $f(x)$.

2 Answers2