1194. Find the power series expansion of the particular integral of the equation $y^{\prime \prime}+x y=0$, satisfying the initial conditions: $y(0)=1, y^{\prime}(0)=0$.
Solution. Let the desired function $y(x)$ be expanded into a Maclaurin series

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

where the quantities $y(0), y^{\prime}(0), y^{\prime \prime}(0), \ldots$ are the values of the function $y(x)$ and its derivatives at $x=0$.

The first two coefficients $y(0)$ and $y^{\prime}(0)$ are given in the problem statement, the third is obtained by substituting the known values into the given equation, $y^{\prime \prime}(0)=0$, and the subsequent coefficients are found by successive differentiation of the given equation: $y^{\prime \prime \prime}=-\left(y+x y^{\prime}\right) ; \quad y^{(4)}=-\left(2 y^{\prime}+x y^{\prime \prime}\right) ; \quad y^{(5)}=-\left(3 y^{\prime \prime}+x y^{\prime \prime \prime}\right) ; \ldots ;$ $y^{(n)}=-\left[(n-2) y^{(n-3)}+x y^{(n-2)}\right] ; \ldots$ From this, at $x=0$ we obtain: $y^{\prime \prime \prime}(0)=-1 ; y^{(4)}(0)=y^{(6)}(0)=0 ; y^{(6)}(0)=1 \cdot 4 ; y^{(7)}(0)=y^{(8)}(0)=0$; $y^{(\theta)}(0)=-1 \cdot 4 \cdot 7 ; \ldots$

Substituting these values of the coefficients into the Maclaurin series (2), we obtain the desired particular integral in the form of a series

$y=1-\frac{1}{3!} x^{3}+\frac{1 \cdot 4}{6!} x^{6}-\frac{1 \cdot 4 \cdot 7}{9!} x^{3}+\ldots+(-1)^{m} \frac{1 \cdot 4 \cdot 7 \ldots(3 m-2)}{(3 m)!} x^{3 m}+\ldots$, which converges for any value of $x$.