Find the differentiable function $f$ such that $f(x) = x^2+\int_0^{x} e^{-t }f(x−t)dt$ $f$ is a differentiable function such that $f(x) = x^2+\int_0^{x} e^{-t} f(x−t)dt$ then find $f(x)$
I had used Newton Leibniz partial differentiation and substitution of $x$ to $x+t$ but I am not able to get anyone solve it. You can use Newton Leibniz rule it will help I think 
 A: Let $g(x) = e^xf(x)$. Multiplying the equation for $f$ with $e^x$, you can see that
$$
g(x) = x^2e^x + \int_0^xg(s) ds
$$ 
which implies
$$
g'(x) = (2x+x^2)e^x + g(x)  
$$ 
with $g(0) = 0$. Therefore 
$$
\frac{d}{dx} (e^{-x} g(x)) = \frac{d}{dx} f(x) = 2x + x^2
$$
implying 
$$
f(x) = x^2 + \frac{1}{3}x^3.
$$
A: The initial condition is $f(0) = 0$. We now have all we need to take a Laplace transform in the variable $x$. Before that, we need the following:

Theorem. (Convolution) Define the convolution of two functions $f$ and $g$ by
  $$ (f * g)(x) := \int_0 ^x f(t)g(x-t) \ dt $$
  Then the Laplace transform of the convolution is
  $$ \mathcal{L}(f * g) = F(s)G(s) $$
  where $F$ and $G$ are the transforms of $f$ and $g$ respectively.

Let's get started. I'll assume you know some basic Laplace transforms (these can be easily searched). First take transforms on both sides.
$$ F(s) = \frac{2}{s^3} + \frac{F(s)}{s-1} $$
Now move some things around.
$$ F(s) \left( 1 - \frac{1}{s-1} \right) = \frac{2}{s^3} $$
And keep going...
$$ F(s) = \frac{2(s-2)}{(s-1)s^3} $$
So, without inverting (yet, I'll leave this to you via partial fractions and some table matching) the solution looks like something with some $x^2$ terms and perhaps even $e^{-2x}$. Of course, this is just one way to solve the problem - my favorite actually!
