$f(x)$ is a differentiable function satisfying the relation $f(x)=x^2+\int_0^x e^{-t} f(x-t) dt$, then $\sum\limits_{k=1}^9f(k)$ is equal to

Question

$f(x)$ is a differentiable function satisfying the relation $f(x)=x^2+\int_0^x e^{-t} f(x-t) dt$, then $\sum\limits_{k=1}^9f(k)$ is equal to

I first differentiated the given relation on both sides w.r.t. $x$ and obtained $f'(x)=2x+e^{-x} f(0).$ Substituting $0$ in the given relation, we have $f(0) = 0$. By substituting the value we get $f'(x)=2x$. Now integrating this on both sides would result in $f(x)=x^2+c$; again for $x=0$ we get $c = 0$. Thus $f(x)=x^2$. Turns out the answer is $$f(x)=x^2+\frac{x^3}{3}.$$ Could someone please explain where I went wrong?

Answer 1

Let $v=x-t,$ $$\int_{0}^{x}e^{-t}f(x-t)dt=e^{-x}\int_{0}^{x}e^{v}f(v)dv$$

Differentiate the equation on both sides w.r.t $x$ gives:$$f'(x)=2x+f(x)-e^{-x}\int_{0}^{x}e^{v}f(v)dv=2x+f(x)-f(x)+x^2$$ $$f'(x)=2x+x^2$$ Then solve the ODE with initial condition $f(0)=0$.

Answer 2

Before differentiating use the IVth property of definite integration that $$\int_{0}^{a} g(t) dt=\int_{0}^{a} g(a-t) dt.$$ Then $$\int_{0}^{x} e^{-t} f(x-t) dt=e^{-x}\int_{0}^{x} e^{t} f(t) dt.$$ Next you get $$f(x)=x^2+e^{-x}\int_{0}^{x}e^{t} f(t) dt.~~~~~~(*)$$ Note that $f(0)=0$.Now differentiate w.r.t.x $$f'(x)=2x-e^{-x}\int_0^{x} e^{t}f(t) dt+e^{-x} e^xf(x)$$ Next, use (*) to write $$f'(x)=2x-f(x)-x^2+f(x)\implies f'(x)=2x+x^2\implies f(x)=x^2+x^3/3+C$$ $f(0)=0$ gives $C=0$.

Answer 3

Here is an alternative solution using the Laplace transform. Taking the transform of both sides of the equation, we get $$ \mathcal{L}\{f(x)\}=\mathcal{L}\{x^2\}+\mathcal{L}\{e^{-x}\}\mathcal{L}\{f(x)\} =\frac{2}{s^3}+\frac{1}{s+1}\,\mathcal{L}\{f(x)\} $$ $$ \implies\mathcal{L}\{f(x)\}=\frac{2}{s^3}+\frac{2}{s^4} \implies f(x)=x^2+\frac{x^3}{3}. $$