1195. Find the first five terms of the power series expansion of the particular integral of the equation $y^{\prime \prime}-y e^{x}=0$, satisfying the initial conditions: $y(0)=2, y^{\prime}(0)=1$.
Solution. Applying the same method as in the solution of the previous problem, we get:

$$
\begin{aligned}
& y^{\prime \prime}=y e^{x}, \quad y^{\prime \prime}(0)=2 \\
& y^{(3)}=\left(y+y^{\prime}\right) e^{x}, \quad y^{(3)}(0)=3 \\
& y^{(4)}=\left(y+2 y^{\prime}+y^{\prime \prime}\right) e^{x}, \quad y^{(4)}(0)=6
\end{aligned}
$$

$$
y=2+x+x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{4}+\ldots
$$

This second method for determining the coefficients of the power series that satisfies the given differential equation, which is based on the use of the Maclaurin series*, in some cases requires less computational work than the method of undetermined coefficients. It is applicable for finding the general or particular integrals of the equation if it is solvable with respect to the highest-order derivative and if it is possible to obtain a derivative of any order by its successive differentiation.

Integration of equations using series is of great importance, however, it should be noted that not every equation can yield an integral in the form of a suitable power series.

For example, the equation $x^{2} y^{\prime}-y(x+1)=-x^{2}$ (linear) has the general integral

$$
y=\frac{x}{\sqrt[x]{e}}\left(c-\int \frac{\sqrt[x]{e}}{x} d x\right)
$$

However, assuming that there exists an integral in the form of a power series (1) and determining its coefficients, we obtain the series

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

which is practically unusable, as it diverges for any value of $x$ different from zero.