Consider the integral equation
$$ y(x) = x + \int_0^x (t-x)y(t) \, dt, \quad x \in [0,1].$$
My goal is to use the iterative scheme of succesive approximations to find the unique solution of the equation above. As a suggestion, my lecturer told us to consider the initial approximation $y_0(x) = x.$
My attempt. Considering $y_0(x) = x,$ it follows that $$ y_1(x) = x + \int_0^x (t-x)t\, dt = x + \int_0^x t^2 -xt \, dt = x - \frac{x^3}{3!}.$$ $$ y_2(x) = x + \int_0^x (t-x)(t - \frac{t^3}{3!}) \, dt = x + \int_0^x t^2 - \frac{t^4}{3!}-xt+x\frac{t^3}{3!} \, dt = x - \frac{x^3}{3} + \frac{x^5}{5!}.$$ This leads me to conclude that $$y_n(x) = \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!}x^{2k+1}.$$ Then, $$ \lim_{n \rightarrow \infty} y_n(x) = \sin(x).$$ Note that the series converges for all $x$, in particular it also converges for $x \in [0,1].$
My concerns. Is my resolution formal enough? How would I be able to prove the expression for $y_n(x)?$
As for the second question, follows my attempt (not succesfull): We want to show that $$ y_n(x) = \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!}x^{2k+1}, \quad \forall n \geqslant 0.$$ The base cases $n=0,1,2$ are succesfully verified with my calculations above. Now suppose that the equality is true for some integer $n \in \mathbb N.$ Then, \begin{equation} \begin{split} y_{n+1}(x) &= x + \int_0^x (t-x)y_n(t) \, dt = x + \int_0^x (t-x)\left(\sum_{k=0}^n \frac{(-1)^k}{(2k+1)!} t^{2k+1}\right) \, dt \\[.2cm] &= x + \int_0^x \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!}t^{2k+2} \, dt - x\int_0^x \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!}t^{2k+1} \, dt \\[.2cm] &= x + \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!}\frac{x^{2k+3}}{(2k+3)} - \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!} \frac{x^{2k+2}}{(2k+2)}. \end{split} \end{equation}
I don't see how to get to the result from here.
Thanks for any help in advance.
Error in last line: $x^{2k+2}$ should be $x^{2k+3}$. Also $\frac{1}{2k+3}-\frac{1}{2k+2}=\frac{-1}{(2k+2)(2k+3)}$ giving the result you want after adjusting index.