Given $\frac{dy}{dx}=x^2+y^2$ and initial condition $\varphi (0)=1$, find the first 6 terms in the Taylor expansion solution $y=\varphi (x)$ Given $\frac{dy}{dx}=x^2+y^2$ and initial condition $\varphi (0)=1$, use the method of reduction to an integral equation and successive approximation to find the first 6 terms in the Taylor expansion solution $y=\varphi (x)$.
We have that if $f:D \rightarrow \mathbb{R}$ is continuous, $\varphi$ is defined and continuous on $I=\{x|x_0-h<x<x_0+h\}$ to $\mathbb{R}$, and $(x_0,y_0) \in D$ with $\varphi (x_0)=y_0$, then $\varphi$ is a solution of $\frac{d\varphi}{dx}=f[x,\varphi (x)]$ on $I$ only if $\varphi (x)=y_0+\int_{x_0}^{x} f[t,\varphi (t)] dt$ for $x \in I$. This is what I mean by reduction to an integral equation.
EDIT:
In case it helps anyone refine their answer, the answer should be $1+x+x^2+\frac{4}{3}x^3+\frac{7}{6}x^4+\frac{6}{5}x^5$. This comes out of the back of my book. Of course, my problem is that I cannot get to this answer.

$$y = 1 + \int_0^x y^2 + t^2 \, dt$$  we will define $$y_{n+1} = 1 + \int_0^x (t^2 + y_n^2)\, dt = 1 + \frac 13 x^3 + \int_0^x y_n^2 \, dt,\,  y_0 = 1.$$   so $$y_1 = 1+ \frac 13 x^3 + \int_0^x  \,dt = 1 + x + \frac 13 x^3 \tag 1\\ y_2 = 1 + \frac 13 x^3 + \int_0^x \left(1 + t + \frac 13 t^3\right)^2\, dt = 1 + \frac 13 x^3 + \int_0^x  \left(1 + 2t + t^2 + \frac 23 t^3 + \frac 23 t^4 + \frac 19 t^6\right)$$
$$y_2 = 1 + x+ x^2 + \frac23 x^3+\frac 1{6}x^4+\frac 2{15}x^5+\frac 1{63}x^7\tag 2\\
y_2^2 = 1 + 2x + 3x^2 + \frac{10} 3 x^3+\frac 83 x^4 + \cdots\\
y_3 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{10} 3 t^3+\frac 83 t^4 + \cdots  \right)dt $$
$$y_3=1 + x+x^2 + \frac 43 x^3+\frac 5{17}x^4 + \frac 8{15}x^5+\cdots\tag 3\\
y_3^2 = 1 + 2x + 3x^2 + \frac{14}3x^3+\cdots$$
$$y_4 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{14} 3 t^3+ \cdots  \right)dt$$
$$
\\y_4 = 1 + x + x^2 + \frac43x^3 + \frac 76 x^4 + \cdots \tag 4
$$
