Solve ODE $x'=x^2$ using Power Series I need to solve $x'=x^2$, $x(0)=1$ using the power series $\sum_{n=0}^{\infty}a_n t^{n}$ and show that it has a solution on $(-1,1)$ that can be extended to $(-\infty,1)$.
I have rewritten the ODE as $\sum_{n=1}^{\infty}{na_n t^{n-1}}-\sum_{n=0}^{\infty}a_n t^{2n}=0$ and then as $\sum_{n=0}^{\infty}{(n+1)a_n t^{n}}-\sum_{n=0}^{\infty}a_n t^{2n}=0$.
Here I don't know how to continue, I would be very grateful for a step by step solution, because I still need to solve more problems like this one.
 A: *

*If you suppose that there exists a solution in power series of the form $y=\sum_{n=0}^{+\infty}a_n t^n$, then $y'=\sum_{n=1}^{+\infty}na_{n}t^{n-1}=\sum_{n=0}^{+\infty}(n+1)a_{n+1}t^n$


*By Cauchy product $y^2=(\sum_{n=0}^{+\infty}a_nt^n)\cdot (\sum_{n=0}^{+\infty}a_nt^{n})=\sum_{n=0}^{+\infty}c_nt^{n}$ with $c_n=\sum_{k=0}^{n}a_{k}a_{n-k}$.


*Thus, $x'=x^2$ can be written as $\sum_{n=0}^{+\infty}(n+1)a_{n+1}t^{n}=\sum_{n=0}^{+\infty}c_n t^n$ and so $c_n=(n+1)a_{n+1}$.


*Equality  coefficients: If $n=0$, then $c_0=a_0^2$ but then $a_1=a_0^2$. If $n=1$, then $c_1=a_0a_1+a_1a_0=2a_0a_1$ but then $2a_0a_1=2a_2$. If $n=2$, then $c_2=a_0a_2+a_1a_1+a_2a_0=2a_0a_2+a_1^2$ but then $3a_3=2a_0a_1+a_1^2$.
Try to guess a formula for $a_n$ inductively and then use the initial condition for to find the particular solution.
A: You made a mistake in replacing $x^2$ with its series form. Looks like all you did was swap out $t^n$ for $t^{2n}$. There's much more to it that makes solving the ODE with this method more complicated.
Assuming a series solution
$$x(t) = \sum_{n\ge0} a_n t^n$$
its square, using the multinomial theorem, would be
$$x(t)^2 = \left(\sum_{n\ge0} a_n t^n\right)^2 = \sum_{n\ge0} {a_n}^2 t^{2n} + 2 \sum_{1\le m<n<\infty} a_m a_n t^{m+n}$$
Or, as a Cauchy product, we have
$$x(t)^2 = \sum_{m\ge0} a_m t^m \cdot \sum_{n\ge0} a_n t^n = \sum_{m\ge0} \sum_{0\le n\le m} a_n a_{m-n} t^m$$
Neither of these look like they play nice.
Hint: Instead of trying to attack right away with series, make a substitution to get a linear ODE, which is much more amenable to the series method.
