Example 4. Find a particular solution of the differential equation

$$
x^{4} \cdot y^{\prime \prime \prime}-2 x^{3} \cdot y^{\prime \prime}=\frac{5}{x}
$$

satisfying the initial conditions $y(1)=-\frac{1}{18}, y^{\prime}(1)=-\frac{7}{18}$, $y^{\prime \prime}(1)=\frac{1}{6}$.
Solution. The given equation belongs to type 2) with $n=3$ and $k=2$. Let $y^{\prime \prime}=p(x)$. Then $y^{\prime \prime \prime}=p^{\prime}$. After dividing the equation term by term by $x^{4}$ and making the appropriate substitutions, we arrive at the equation

$$
p^{\prime}-\frac{2}{x} p=\frac{5}{x^{5}}
$$

This is a first-order linear equation, which we solve by the substitution $p=u v, p^{\prime}=u^{\prime} v+v u^{\prime}$. Omitting the intermediate calculations (which need to be performed independently), we indicate that the general solution of this linear equation is

$$
p=x^{2}\left(C_{1}-\frac{5}{6 x^{6}}\right), \quad \text { i.e. } \quad y^{\prime \prime}=x^{2}\left(C_{1}-\frac{5}{6 x^{6}}\right)
$$

After two integrations, we find the general solution of the original equation:

$$
\begin{gathered}
y^{\prime}=\int\left(C_{1} x^{2}-\frac{5}{6 x^{4}}\right) d x=\frac{C_{1} x^{3}}{3}+\frac{5}{18 x^{3}}+C_{2} \\
y_{\text {general }}=\frac{C_{1} x^{4}}{12}-\frac{5}{36 x^{2}}+C_{2} x+C_{3}
\end{gathered}
$$

The values of the three constants will be determined based on the initial conditions:

$$
x=1, \quad y=-\frac{1}{18}, \quad y^{\prime}=-\frac{7}{18}, \quad y^{\prime \prime}=\frac{1}{6}
$$

We start with the equation $y^{\prime \prime}=x^{2}\left(C_{1}-\frac{5}{6 x^{6}}\right)$, into which we substitute $x=1$, $y^{\prime \prime}=\frac{1}{6}: \frac{1}{6}=1\left(C_{1}-\frac{5}{6}\right)$. We find $C_{1}=1$. Now, in the equation $y^{\prime}=\frac{C_{1} x^{3}}{3}+\frac{5}{18 x^{3}}+C_{2}$, we substitute: $x=1, y^{\prime}=-\frac{7}{18}, C_{1}=1$. We get $C_{2}=-1$. Finally, in the equation $y=\frac{C_{1} x^{4}}{12}-\frac{5}{36 x^{2}}+C_{2} x+C_{3}$, we substitute $x=1, \quad y=-\frac{1}{18}, \quad C_{1}=1, \quad C_{2}=-1: \quad-\frac{1}{18}=\frac{1}{12}-\frac{5}{36}-1+C_{3}$. We find $C_{3}=1$.

$$
\text { Answer. } y_{4}=\frac{x^{4}}{12}-\frac{5}{36 x^{2}}-x+1 \text {. }
$$