1191. Find the general integral of the equation $\frac{d y}{d x}=y^{2}$ in the form of a power series.
Solution. Let the desired integral be a power series

$$
y=a_{11}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}+\ldots
$$

where $a_{0}, a_{1}, \ldots, a_{n}, \ldots$ are unknown constants to be determined.

Assuming that such a series exists and converges in some interval of $x$ values, we find the series for $\frac{d y}{d x}$ by term-by-term differentiation

$$
\frac{d y}{d x}=a_{1}+2 a_{2} x+3 a_{3} x^{2}+\ldots+n a_{n} x^{n-1}+\ldots
$$

and the series for $y^{2}$ by term-by-term multiplication of the series (1) with itself:

$$
y^{2}=a_{0}^{2}+2 a_{0} a_{1} x+2 a_{0} a_{2} x^{2}+2 a_{11} a_{3} x^{3}+\ldots+a_{1}^{2} x^{2}+2 a_{1} a_{2} x^{3}+\ldots
$$

Substituting these series for $\frac{d y}{d x}$ and $y^{2}$ into the given equation and equating the coefficients of like powers of $x$ from both sides, since two series will be identically equal only under this condition, we obtain the following system:

$$
\begin{array}{l|l}
x^{0} & a_{1}=a_{0}^{2} \\
x^{1} & 2 a_{2}=2 a_{0} a_{1} \\
x^{2} & 3 a_{3}=2 a_{0} a_{2}+a_{1}^{2} \\
x^{3} & 4 a_{4}=2 a_{0} a_{3}+2 a_{1} a_{2}
\end{array}
$$

Solving this system, we find: $a_{1}=a_{0}^{2} ; a_{2}=a_{0}^{3} ; a_{3}=a_{0}^{4} ; \ldots$ : $a_{n}=a_{0}^{n+1} ; \ldots$

Therefore, the desired power series expansion of the general integral of the given equation is

$$
y=a_{0}\left(1+a_{0} x+a_{0}^{2} x^{2}+a_{0}^{3} x^{3}+\ldots+a_{0}^{n} x^{n}+\ldots\right)
$$

where $a_{0}$ is an arbitrary constant.

The obtained series represents an infinite geometric progression with common ratio $q=a_{0} x$ and for $|q|<1$ has the sum $y=\frac{a_{0}}{1-a_{0} x}$.

The solution of this problem demonstrates the reliability of the method of integrating equations using series, as direct integration of the given equation, as a first-order equation with separable variables, yields the same result.