Find exact solution of Cauchy Problem using the convergence of a sequence defined by Fixed Point Theorem I have the following Cauchy problem 
$$
\begin{cases}
y'= y^2 - x^2= f(x,y)\\
y(0)=1
\end{cases}
$$
in the rectangle $R=\{(x,y), 0 \leq x \leq 1, |y-1| \leq 1\}$.
The question is to prove the existence and uniqueness of solution of the problem and find the exact solution or an approximate solution.
Using the fixed point Theorem, I proved that the Cauchy problem admits a unique solution in the interval $\big[0,\frac{1}{5}\big]$.
For calculation of the solution, we have by the Fixed Point Theorem that the sequence $(y_n)= T(y_{n-1})$ converge vers the unique solution of the problem. We have:
$$
T\big(y(x)\big)= y_0+ \int_0^x (y^2(s)-s^2)\ ds, \ y \in R, \ x \in \left[0,\dfrac{1}{5}\right]
$$
Then by the relation $y_n= T\big(y_{n-1}(x)\big)$, we have 
\begin{align}
y_1(x) &= 1+ \int_0^x (1-s^2) ds = 1 + x-\dfrac{x^3}{3} \\
y_2(x) &= 1+ x +x^2 -\dfrac{1}{6} x^4 - \dfrac{2}{16} x^5 + \dfrac{1}{63} x^7
\end{align}
How do we find the exact solution $y(x)$?
Thanks in advance for the help.
 A: Your equation is a Riccati equation, you can not find a simple symbolic solution. However, you can insert the power series, assuming it exists and converges, either into the differential equation or into the integral version.
$$
y=\sum_{n=0}^\infty c_nx^n=1-\frac{x^3}3+\sum_{n=1}^\infty \frac{x^n}{n}\sum_{k+m=n-1}c_kc_m
$$
so that by comparing coefficients you get
\begin{align}
c_0&=1\\
c_1&=c_0^2=1\\
c_2&=\frac12(c_0c_1+c_1c_0)=1\\
c_3&=-\frac13+\frac13(c_1^2+2c_0c_2)=\frac23\\
c_4&=\frac12(c_0c_3+c_1c_2)=\frac56\\
&\vdots\\
c_{2m}&=\frac1m(c_0c_{n-1}+c_1c_{n-2}+\dots+c_{m-1}c_m)\\
c_{2m+1}&=\frac1{2m+1}(2c_0c_{2m}+2c_1c_{2m-1}+\dots-2c_{m-1}c_{m+1}+c_m^2)
\end{align}

A systematic way to solve it is to set $y=-\frac{u'}u$ so that the second order linear ODE
$$
u''=x^2u, ~ u(0)=1,~u'(0)=-1 
$$
results. This is easier to solve via power series expansion as a simple coefficient recursion results,
$$
a_{n+4}=\frac{a_n}{(n+4)(n+3)},~~ a_0=1,~ a_1=-1,~ a_2=0,~a_3=0
$$
You can express the solution in terms of named special functions like Bessel functions, see Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$ and links there for conversion methods.

You can split off the growth envelope of $u$ as $u(x)=e^{x^2/2}v(x)$ which is the same as setting $y(x)=x-\frac{v'(x)}{v(x)}$. The ODE for $v$ is again order 2 linear,
$$
(x^2+1)v+2xv'+v''=x^2v\\
v''+2xv'+v=0,~~v(0)=1, v'(0)=u'(0)=-1
$$
and we get for the power series expansion
$$
x^{n-2}:n(n-1)b_n +(2(n-2)+1)b_{n-2}=0\\~\\
b_0=1,~~b_1=-1,\\~\\
b_{n}=-\frac{2n-1}{n(n-1)}b_{n-2}\implies
\begin{aligned}
b_{2m}&=(-1)^m\frac{(4m-1)(4m-5)\dots7\cdot 3}{(2m)!}b_0\\~\\
b_{2m+1}&=(-1)^m\frac{(4m+1)(4m-3)\dots9\cdot 5}{(2m+1)!}b_1
\end{aligned}
$$
