Example 2. Solve the equation $x y^{\mathrm{IV}}+y^{\prime \prime \prime}=0$.
Solution. This is an equation of the form (11.2), for which $k=3, n=4$. By taking the lowest derivative $y^{\prime \prime \prime}$ as a new unknown function $u$, we perform the substitution (11.3)

$$
y^{\prime \prime \prime}=u
$$

from which $y^{\mathrm{IV}}=\frac{d u}{d x}$. Thus, the given equation reduces to a first-order equation in terms of $u$:

$$
x u^{\prime}+u=0 \text { or } \frac{d u}{u}+\frac{d x}{x}=0
$$

from which we obtain

$$
u=\frac{C_{1}}{x}
$$

Since $u=y^{\prime \prime \prime}$, to determine $y$ we have the equation

$$
y^{\prime \prime \prime}=\frac{C_{1}}{x}
$$

in which the right-hand side depends only on $x$, i.e., an equation of the form (11.1). Integrating three times, we get respectively:

$$
y^{\prime \prime}=\int \frac{C_{1}}{x} d x=C_{1} \ln x+C_{2}, y^{\prime \prime}=C_{1} \ln x+C_{2}
$$

$$
\begin{gathered}
y^{\prime}=C_{1}(x \ln x-x)+C_{2} x+C_{3} \\
y=C_{1}\left(\frac{x^{2}}{2} \ln x-\frac{3}{4} x^{2}\right)+\frac{C_{2}}{2} x^{2}+C_{3} x+C_{4}
\end{gathered}
$$

or

$$
y=C_{1}^{\prime} x^{2} \ln x+C_{2}^{\prime} x^{2}+C_{3} x+C_{4},
$$

where

$$
C_{1}^{\prime}=\frac{C_{1}}{2}, C_{2}^{\prime}=\frac{C_{2}}{2}-\frac{3}{4} C_{1}
$$