Example 7. Solve the equation $x^{3} y^{\prime \prime}=\left(y-x y^{\prime}\right)^{2}$.
Solution. We will show that this equation is a generalized homogeneous equation. Considering $x, y, y^{\prime}, y^{\prime \prime}$ as quantities of the 1st, $m$-th, $(m-1)$-th, and $(m-2)$-th dimensions respectively, and equating the dimensions of all terms, we get

$$
3+(m-2)=2 m
$$

from which $m=1$. The solvability of equation (2) is a condition for the generalized homogeneity of the equation.

We make the substitution $x=e^{t}, y=u e^{t}$. Since

$$
\begin{aligned}
& \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{(d u / d t+u) e^{t}}{e^{t}}=\frac{d u}{d t}+u \\
& \frac{d^{2} y}{d x^{2}}=\frac{(d / d t)(d y / d t)}{d x / d t}=\frac{d^{2} u / d t^{2}+d u / d t}{e^{t}}=e^{-t}\left(\frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}\right)
\end{aligned}
$$

the given equation, after canceling the factor $e^{2 t}$, will take the form

$$
\frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}=\left(\frac{d u}{d t}\right)^{2}
$$

Setting $\frac{d u}{d t}=p, \frac{d^{2} u}{d t^{2}}=p \frac{d p}{d u}$, we get $p \frac{d p}{d u}+p=p^{2}$. Hence, $p=0$ or $\frac{d p}{d u}+1=p$. Integrating the second equation, we find

$$
p=1+C_{1} e^{u}, \quad \text { or } \quad \frac{d u}{d t}=1+C_{1} e^{u}
$$

The general solution of this equation is

$$
u=\ln \frac{e^{t}}{C_{1} e^{t}+C_{2}}
$$

Returning to the variables $\boldsymbol{x}$ and $\boldsymbol{y}$, we obtain the general solution of the given equation

$$
y=x \ln \frac{x}{C_{1} x+C_{2}}
$$

The case $p=0$ gives $u=C$ or $y=C x$ - a particular solution, which is obtained from the general solution when $C_{1}=e^{-c} ; C_{2}=0$.

Remark. When solving the Cauchy problem for higher-order equations, it is advisable to determine the values of the constants $C_{i}$ during the solution process, rather than after finding the general solution of the equation. This speeds up the solution of the problem and, moreover, it may turn out that integration is significantly simplified when the constants $C_{i}$ take specific numerical values, while for arbitrary $C_{i}$, integration may be difficult or even impossible in elementary functions.