$f(x)$ 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. $x$; $f'(x)=2x+e^(-x) f(0)$ . 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 wold 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?

$f(x)$ 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?

$f(x)$ 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. $x$; there would be a $f(0)$ which would be equal to $0$. I substituted the value and integrated on both sides leading to the answer $f(x)=x^2$. Turns out $$f(x)=x^2+\frac{x^3}{3}.$$ Could someone please explain where I went wrong?

$f(x)$ 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. $x$; $f'(x)=2x+e^(-x) f(0)$ . 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 wold 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?

Question including the relation

I first differentiated the given relation on both sides w.r.t. $x$; there would be a $f(0)$ which would be equal to $0$. I substituted the value and integrated on both sides leading to the answer $f(x)=x^2$. Turns out $$f(x)=x^2+\frac{x^3}{3}.$$ Could someone please explain where I went wrong?

$f(x)$ 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. $x$; there would be a $f(0)$ which would be equal to $0$. I substituted the value and integrated on both sides leading to the answer $f(x)=x^2$. Turns out $$f(x)=x^2+\frac{x^3}{3}.$$ Could someone please explain where I went wrong?