Example 3. Find the first four terms of the Taylor series expansion of the solution $y=y(x)$ of the equation $y^{\prime \prime}=e^{x y}$, satisfying the initial conditions $\left.y\right|_{x=0}=1,\left.y^{\prime}\right|_{x=0}=0$.
Solution. It is easy to see that the right-hand side of the equation, i.e., the function $e^{z y}$, can be expanded into a power series in powers of $x$ and $y$ in a neighborhood of the point (0,0), converging in the domain $-\infty < x < +\infty, -\infty < y < +\infty$ (i.e., the right-hand side is holomorphic).

We will seek a particular solution in the form of a series

$$
y(x) = y(0) + \frac{y'(0)}{1!} x + \frac{y''(0)}{2!} x^2 + \frac{y'''(0)}{3!} x^3 + \ldots
$$

Using the equation itself, we find $y''(0) = \left. e^{x y} \right|_{z=0} = 1$.

Differentiating both sides of the equation successively and setting $x=0$ in the resulting equalities, we will have

$$
\begin{aligned}
y'''(0) & = \left. \left( y + x y' \right) e^{x y} \right|_{x=0} = 1 \\
y^{(4)}(0) & = \left. \left[ 2 y' + x y'' + \left( y + x y' \right)^2 \right] e^{x y} \right|_{x=0} = 1.
\end{aligned}
$$

Substituting the found values of $y''(0), y'''(0), y^{(4)}(0)$ into the series (20), we obtain the desired expansion of the solution

$$
y(x) = 1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots
$$

## Problems for Independent Solution

In the following problems, find the first three terms of the power series expansion of the solution of the given differential equation for the given initial conditions:

724. $y' = 1 - x y$,

$\left. y \right|_{x=0} = 0$.

725. $y' = \frac{y - x}{y + x}$

$\left. y \right|_{x=0} = 1$.

726. $y' = \sin x y$,

$\left. y \right|_{x=0} = 1$.

727. $y'' + x y = 0$,

$\left. y \right|_{x=0} = 0, \left. y' \right|_{x=0} = 1$.

728. $y'' - \sin x y' = 0$,

$\left. y \right|_{x=0} = 0, \left. y' \right|_{x=0} = 1$.

729. $x y'' + y \sin x = x$,

$\left. y \right|_{x=\pi} = 1, \left. y' \right|_{x=\pi} = 0$.

730. $y'' \ln x - \sin x y = 0$;

$\left. y \right|_{x=e} = e^{-1}, \left. y' \right|_{x=e} = 0$.

731. $y''' + x \sin y = 0$,

$\left. y \right|_{x=0} = \frac{\pi}{2}, \left. y' \right|_{x=0} = 0, \left. y'' \right|_{x=0} = 0$.

Integrate the following differential equations using series:

732. $y' - 2 x y = 0$,

$y(0) = 1$.

733. $y'' + x y' + y = 0$.
734. $y'' - x y' + y - 1 = 0, \quad y(0) = y'(0) = 0$.

In problems 735-738, find six terms of the expansion of $y(x)$:

735. $y'' - (1 + x^2) y = 0, \quad y(0) = -2, \quad y'(0) = 2$.
736. $y'' = x^2 y - y', \quad y(0) = 1, y'(0) = 0$.
737. $y'' - y e^x = 0$.
738. $y' = e^y + x y, \quad y(0) = 0$.

## $2^{\circ}$. Expansion of the Solution in a Generalized Power Series.

The Bessel Equation. A point $x_0$ is called an ordinary point of the differential equation

$$
y'' + p(x) y' + q(x) y = 0
$$

if the coefficients $p(x)$ and $q(x)$ are holomorphic at this point; otherwise, the point $x_0$ is called a singular point of the differential equation (21).

A series of the form

$$
x^{\rho} \sum_{k=0}^{\infty} c_k x^k \quad (c_0 \neq 0)
$$

where $\rho$ is a given number, and the power series $\sum_{k=0}^{\infty} c_k x^k$ converges in some domain $|x| < R$, is called a generalized power series.

If $\rho$ is a non-negative integer, then the generalized power series (22) becomes an ordinary power series.

Theorem. If the point $x=0$ is a singular point of the equation (21), and the coefficients $p(x)$ and $q(x)$ of the equation can be represented as

$$
p(x) = \frac{\sum_{k=0}^{\infty} a_k x^k}{x}, \quad q(x) = \frac{\sum_{k=0}^{\infty} b_k x^k}{x^2}
$$

where the series in the numerators converge in some domain $|x| < R$, and the coefficients $a_0, b_0$, and $b_1$ are not all zero simultaneously, then the equation (21) has at least one solution in the form of a generalized power series

$$
y = x^{\rho} \sum_{k=0}^{\infty} c_k x^k \quad (c_0 \neq 0)
$$

which converges, at least, in the same domain $|x| < R$.

To determine the exponent $\rho$ and the coefficients $c_k$, one needs to substitute the series (22) into the equation (21), cancel by $x^{\rho}$, and set the coefficients of all powers of $x$ to zero (the method of undetermined coefficients).

The number $\rho$ is found from the so-called indicial equation

$$
\rho(\rho-1) + a_0 \rho + b_0 = 0
$$

where

$$
a_0 = \lim_{x \to 0} x p(x), \quad b_0 = \lim_{x \to 0} x^2 q(x)
$$

Let $\rho_1$ and $\rho_2$ be the roots of the indicial equation (23). We will consider three cases.

1. If the difference $\rho_1 - \rho_2$ is not an integer or zero, then two solutions of the form (22) can be constructed

$$
y_1 = x^{\rho_1} \sum_{k=0}^{\infty} c_k x^k \quad (c_0 \neq 0), \quad y_2 = x^{\rho_2} \sum_{k=0}^{\infty} A_k x^k \quad (A_0 \neq 0)
$$

2. If the difference $\rho_1 - \rho_2$ is a positive integer, then, generally speaking, only one series (solution of equation (21)) can be constructed

$$
y_1 = x^{\rho_1} \sum_{k=0}^{\infty} c_k x^k
$$

3. If the indicial equation (23) has a multiple root $\rho_1 = \rho_2$, then only one series - the solution (25) can be constructed. It is clear that in the first case, the constructed solutions $y_1(x)$ and $y_2(x)$ will be linearly independent (i.e., their ratio will not be a constant).

In the second and third cases, we have constructed only one solution. Note that if the difference $\rho_1 - \rho_2$ is a positive integer or zero, then in addition to the solution (25), the equation (21) will have a solution of the form

$$
y_2 = A y_1(x) \ln x + x^{\rho_2} \sum_{k=0}^{\infty} A_k x^k
$$

In this case, $y_2(x)$ contains an additional term of the form $A y_1(x) \ln x$, where $y_1(x)$ is given by (25).

Remark. The constant $A$ in (26) may turn out to be zero, and then for $y_2$ we will obtain an expression in the form of a generalized power series.