Series Solution of Nonlinear ODE I am trying to solve for the series solution of the following nonlinear ODE: $$y'+y^2=0$$ I am stuck in the following step: $$\sum_{n=1}^\infty a_nnx^{n-1}+\sum_{n=0}^\infty \left ( \sum_{k=0}^n a_ka_{n-k} \right )x^n=0$$ Is it possible to solve the DE this way?
 A: Shift the index for the term on the left, so it's $\sum_{n=0}^\infty (n+1) a_{n+1} x^n$.
The recursion $(n+1) a_{n+1} + \sum_{k=0}^n a_k a_{n-k} = 0$ doesn't look promising, but if you look at the first few terms you'll see a pattern.
A: Why insisting on a series solution. Just rewrite the equation
$$\frac{y'}{y^2}+1=-\left(\frac{1}{y}\right)'+1=0$$
an integral of which is of the form $y(x)=\frac{1}{x+C}$ where $C$ is a constant and then if you need a series just develop as
$$\frac{1}{C}\frac{1}{1+\frac{x}{C}}=\frac{1}{C}\sum_{k=0}^\infty(-1)^k(\frac{x}{C})^k$$
A: It might be more efficient to attack the problem with force if the general pattern is not readily seen.
Let
\begin{align}
y(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n} 
\end{align}
for which the equation $y' + y^{2} = 0$ leads to
\begin{align}
a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + \cdots &= - (a_{0} + a_{1} x + a_{2} x^{2} + \cdots )^{2} \\
&= - [ a_{0}^{2} + (a_{0}a_{1} + a_{1} a_{0}) x + \cdots ]. 
\end{align}
Equating the coefficients leads to
\begin{align}
a_{1} &= - a_{0}^{2} \\
2 a_{2} &= - (a_{0}a_{1} + a_{1} a_{0}) 
\end{align}
and so on. What is then determined is the general form of the coefficients being
\begin{align}
(n+1) \, a_{n} = - \sum_{k=0}^{n} a_{k} \, a_{n-k}
\end{align}
A: From
$$ \frac{y'}{y^2} = -1 $$
it follows that:
$$ \frac{1}{y(x)} = x+C $$
so:
$$y(x) = \frac{1}{C+x} = \frac{1}{C}-\frac{x}{C^2}+\frac{x^2}{C^3}-\ldots $$
for any $x$ such that $|x|<\frac{1}{|y(0)|}$.
