Example 1. Find the solutions of the equation

$$
y^{\prime \prime}-x y-2 y=0
$$

in the form of a power series.
Solution. We seek $y_{1}(x)$ in the form of a series

$$
y_{1}(x)=\sum_{k=0}^{\infty} c_{k} x^{k}
$$

then

$$
y_{1}^{\prime}(x)=\sum_{k=1}^{\infty} k c_{k} x^{k-1}, \quad y_{1}^{\prime \prime}(x)=\sum_{k=2}^{\infty} k(k-1) c_{k} x^{k-2}
$$

Substituting $y_{1}(x), y_{1}^{\prime}(x)$ and $y_{1}^{\prime \prime}(x)$ into (6), we get

$$
\sum_{k=2}^{\infty}(k-1) k c_{k} x^{k-2}-\sum_{k=1}^{\infty} k c_{k} x^{k}-2 \sum_{k=0}^{\infty} c_{k} x^{k}=0
$$

By combining like terms in (7) and equating the coefficients of all powers of $x$ to zero, we obtain relations from which we can find the coefficients $c_{0}, c_{1}, \ldots$, $c_{n}, \ldots$

Let us assume, for definiteness, that $y_{1}(0)=1, y_{1}^{\prime}(0)=0$. Then it is easy to find that

$$
c_{0}=1, \quad c_{1}=0
$$

Thus, we have

$$
\begin{array}{c|lrl}
x^{0} & 2 c_{2}-2 c_{0}=0, & \text { from this and }(8) & c_{2}=1, \\
x^{1} & 3 \cdot 2 c_{3}-1 \cdot c_{1}-2 c_{1}=0, \text { from this and }(8) & c_{3}=0, \\
x^{2} & 4 \cdot 3 c_{4}-2 c_{2}-2 c_{2}=0, & \text { from this } & c_{4}=\frac{1}{3} \\
x^{3} & 5 \cdot 4 c_{5}-3 c_{3}-2 c_{3}=0, & \text { from this } & c_{5}=0, \\
x^{4} & 6 \cdot 5 c_{6}-4 c_{4}-2 c_{4}=0, & \text { from this } & c_{6}=\frac{c_{4}}{5}=\frac{1}{3 \cdot 5} \\
\ldots & \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{array}
$$

Therefore,

$$
y_{1}(x)=1+x^{2}+\frac{1}{3} x^{4}+\frac{1}{15} x^{6}+\ldots
$$

Similarly, taking

$$
y_{2}(x)=\sum_{k=0}^{\infty} A_{k} x^{k}
$$

and initial conditions $y_{2}(0)=0, y_{2}^{\prime}(0)=1$, we get

$$
A_{0}=0, \quad A_{1}=1
$$

Substituting (10) into (6), we find

$$
\begin{array}{l|l} 
& \sum_{k=0}^{\infty} k(k-1) A_{k} x^{k-2}-\sum_{k=1}^{\infty}(k+2) A_{k} x^{k}=0 \\
x^{0} & 2 A_{2}=0, \quad A_{2}=0 \\
x^{1} & 3 \cdot 2 A_{3}-3 A_{1}=0, A_{3}=\frac{1}{2} \quad(\text { by }(11)), \\
x^{2} & 4 \cdot 3 A_{4}-4 A_{2}=0, A_{4}=0 \\
x^{3} & 5 \cdot 4 A_{5}-5 A_{3}=0, A_{5}=\frac{1}{2 \cdot 4} \\
x^{4} & 6 \cdot 5 A_{6}-6 A_{4}=0, A_{6}=0, \\
x^{5} & 7 \cdot 6 A_{7}-7 A_{5}=0, A_{7}=\frac{1}{2 \cdot 4 \cdot 6}, \ldots \ldots . . \\
\ldots & \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .
\end{array}
$$

It is clear that

$$
A_{2 k}=0, \quad A_{2 k+1}=\frac{1}{2 \cdot 4 \cdot 6 \ldots(2 k)}(k=1,2,3, \ldots)
$$

Thus,

$$
y_{2}(x)=x+\frac{x^{3}}{2}+\frac{x^{5}}{2 \cdot 4}+\frac{x^{7}}{2 \cdot 4 \cdot 6}+\ldots=x \sum_{k=0}^{\infty} \frac{\left(x^{2} / 2\right)^{k}}{k!}=x e^{x^{2} / 2}
$$

The general solution of equation (6) will be of the form

$$
y(x)=A y_{1}(x)+B y_{2}(x)
$$

where $y_{1}(x)$ and $y_{2}(x)$ are given by formulas (9) and (12) respectively, and $A$ and $B$ are arbitrary constants, with $y(0)=A, y^{\prime}(0)=B$.

Let us present another method of integrating differential equations using series, which turns out to be simpler when applied to nonlinear differential equations. Suppose we have a differential equation

$$
y^{(n)}=f\left(x, y, y^{\prime}, \ldots, y^{(n-1)}\right)
$$

and initial conditions

$$
\left.y\right|_{x=x_{0}}=y_{0},\left.\quad y^{\prime}\right|_{x=x_{0}}=y_{0}^{\prime}, \quad \ldots,\left.\quad y^{(n-1)}\right|_{x=x_{0}}=y_{0}^{(n-1)}
$$

We introduce the following definition.

A function $\varphi(x)$ is called holomorphic in some neighborhood $\left|x-x_{0}\right|<\rho$ of the point $x=x_{0}$ if in this neighborhood it can be represented by a power series

$$
\varphi(x)=\sum_{k=0}^{\infty} c_{k}\left(x-x_{0}\right)^{k}
$$

converging in the domain $\left|x-x_{0}\right|<\rho$.

Similarly, a function $\varphi\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is called holomorphic with respect to all its arguments in some neighborhood

$$
\left|x_{k}-x_{k}^{(0)}\right|<\rho_{k} \quad(k=1,2, \ldots, n)
$$

of the point $\left(x_{1}^{(0)}, x_{2}^{(0)}, \ldots, x_{n}^{(0)}\right)$ if it can be represented by a power series

$$
\varphi\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\sum c_{k_{1} k_{2} \ldots k_{n}}\left(x_{1}-x_{1}^{(0)}\right)^{k_{1}}\left(x_{2}-x_{2}^{(0)}\right)^{k_{2}} \ldots\left(x_{n}-x_{n}^{(0)}\right)^{k_{n}}
$$

converging in the domain

$$
\left|x_{k}-x_{k}^{(0)}\right|<\rho_{k} \quad(k=1,2, \ldots, n)
$$

Theorem. If the right-hand side of equation (13) is holomorphic with respect to all its arguments $x, y, y^{\prime}, \ldots, y^{(n-1)}$ in a neighborhood $\Omega$ :

$$
\left|x-x_{0}\right|<R, \quad\left