# 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 $$y_1 = 1+ \frac 13 x^3 + \int_0^x \,dt = 1 + x + \frac 13 x^3 \tag 1\\ y_2 = 1 + \frac 13 x^3 + \int_0^x \left(1 + t + \frac 13 t^3\right)^2\, dt = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + t^2 + \frac 23 t^3 + \frac 23 t^4 + \frac 19 t^6\right)$$ $$y_2 = 1 + x+ x^2 + \frac23 x^3+\frac 1{6}x^4+\frac 2{15}x^5+\frac 1{63}x^7\tag 2\\ y_2^2 = 1 + 2x + 3x^2 + \frac{10} 3 x^3+\frac 83 x^4 + \cdots\\ y_3 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{10} 3 t^3+\frac 83 t^4 + \cdots \right)dt$$ $$y_3=1 + x+x^2 + \frac 43 x^3+\frac 5{17}x^4 + \frac 8{15}x^5+\cdots\tag 3\\ y_3^2 = 1 + 2x + 3x^2 + \frac{14}3x^3+\cdots$$ $$y_4 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{14} 3 t^3+ \cdots \right)dt$$ $$\\y_4 = 1 + x + x^2 + \frac43x^3 + \frac 76 x^4 + \cdots \tag 4$$
• Does not seem to be converging to the real answer. I doubt we can use such technique without guaranteeing the convergence of the sequence $y_{n}$ to the actual solution. – Paramanand Singh Mar 25 '15 at 7:46
The Taylor's series about point $0$ is given by $$\phi(x) = \phi(0) + \phi'(0)x + \frac{\phi''(0)}{2!}x^{2} + \cdots$$ To get six terms we need to calculate derivatives till $\phi^{(6)}(x)$.
Clearly we have $\phi(0) = 1$ and $\dfrac{dy}{dx} = x^{2} + y^{2}$ so that putting $x = 0, y = 1$ we get $\phi'(0) = 0^{2} + 1^{2} = 1$. Again further differentiation gives $$y'' = 2x + 2yy'$$ and so $\phi''(0) = 2\cdot 0 + 2\cdot 1\cdot 1 = 2$. Next we have $$y''' = 2 + 2y'^{2} + 2yy''$$ so that $\phi'''(0) = 2 + 2 + 4 = 8$. Similarly $$y^{(4)} = 6y'y'' + 2yy'''$$ and $\phi^{(4)}(0) = 28$. Again $$y^{(5)} = 6y''^{2} + 8y'y''' + 2yy^{(4)}$$ and $\phi^{(5)}(0) = 144$. Finally $$y^{(6)} = 20y''y''' + 10y'y^{(4)} + 2yy^{(5)}$$ and $\phi^{(6)}(0) = 888$. We then have $$y = \phi(x) = 1 + x + x^{2} + \frac{4}{3}x^{3} + \frac{7}{6}x^{4} + \frac{6}{5}x^{5} + \frac{37}{30}x^{6} + \cdots$$