I want to calculate the limit when $n\rightarrow \infty$of the successive approximation \begin{equation*}y_{n+1}(x)=1+\int_0^xty_n(t)\, dt\end{equation*} with $y_0(x)=1$, $x\in [-1,1]$.
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We have that \begin{align*}&y_0(x)=1 \\ &y_{1}(x)=1+\int_0^xty_0(t)\, dt=1+\int_0^xt\cdot 1\, dt=1+\int_0^xt\, dt=1+\frac{x^2}{2} \\ &y_{2}(x)=1+\int_0^xty_1(t)\, dt=1+\int_0^xt\left (1+\frac{t^2}{2}\right ) \, dt=1+\frac{x^2}{2}+\frac{x^4}{8}\end{align*} Do we have to find a general formula for $y_{n+1}(x)$ or how can we calculate the limit ?
In the limit you are going to get $$y(x)=1+\int_0^xty(t)dt$$ which under differentiation will give the differential equation: $$\frac{dy}{dx}=x\ y(x)\quad,\quad\mbox{y(0)=1}$$ which can be easily solved to get $$y(x)=e^{\frac{x^2}{2}}$$ which upon expanding is $$y(x)=1+\frac{x^2}{2}+\frac{x^4}{8}+\frac{x^6}{48}+...$$