Solution of the integral equation $y(x)+\int_{0}^{x}(x-s)y(s)ds=x^3/6$

So the question is to solve the integral equation to find out $y(x)$.
$$y(x)+\int_{0}^{x}(x-s)y(s)ds=\dfrac{x^3}{6}$$

So I find out the Resolvent kernel for $(x-s)$ and got $\sin{(x-s)}$, so I conclude that the solution is $$y(x)=\dfrac{1}{6}\int_{0}^{x}s^3\sin{(x-s)}ds$$

But the answer says $$y(x)=\int_{0}^{x}s\sin{(x-s)}ds$$

What did I miss?

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In the same idea as Jack D’Aurizio, consider $$A=\int_0^x (x-s) y(s) \, ds$$ from which (using the fundamental theorem of calculus) $$A’=\int_0^x y(s) \, ds$$ $$A”=y(x)$$ So, differentiating twice the original equation $$y”+y=x$$ from which $$y=x+c_1 \cos (x)+c_2 \sin (x)$$

I am sure that you can take from here.

Let $Y(x)=\int_{0}^{x}y(s)\,ds$. By differentiating the original identity with respect to $x$ we get:
$$ y'(x) + Y(x) = \frac{x^2}{2} $$
hence $Y(x) = \frac{x^2-2}{2}+k_1 \cos(x) + k_2\sin(x) $ and:
$$ y(x) = x + k_2\cos(x) -k_1\sin(x).$$
By imposing that the original identity holds, we get $k_2=0$ and $k_1=1$, hence $\color{red}{y(x)=x-\sin(x)}.$

Let

$$Y(p) = \int_0^{\infty} dx \, y(x) \, e^{-p x}$$

be the Laplace transform of $y(x)$. By the convolution theorem, the integral equation is equivalent to

$$Y(p) + \frac1{p^2} Y(p) = \frac1{p^4} $$

Then

$$Y(p) = \frac1{p^2 (p^2+1)} $$

We may find $y(x)$ by finding the inverse LT by either using a table or the residue theorem. Either way,

$$y(x) = x – \sin{x} $$