4.22. Obtain the general solution of the equation $x y^{3} d x+\left(x^{2} y^{2}-\right.$ $-1) d y=0$.
Solution. Let's make the substitution $y=w^{\beta}$, where the parameter $\beta$ will be determined from the condition of equality of the sums of the degrees of each of the monomials standing in parentheses before $d x$ and $d w$: $x w^{3 \beta} d x+\left(x^{2} w^{3 \beta-1}-w^{\beta-1}\right) \beta d w=0$. We equate the sums of the degrees of the monomials: $1+3 \beta=2+3 \beta-1=\beta-1$, from which

$$
\beta=-1, \quad y=\frac{1}{z}
$$

the original equation takes the form of a homogeneous equation: $w x d x+$ $+\left(w^{2}-x^{2}\right) d w=0$. We make the substitution:

$$
\begin{aligned}
& u(x)=\frac{w(x)}{x} \\
& w(x)=x u(x) ; d w=u d x+x d u \\
& u x^{2} d x+\left(u^{2} x^{2}-x^{2}\right)(u d x+x d u)=0 \\
& \left(u x^{2}+u^{3} x^{2}-u x^{2}\right) d x+\left(x^{3} u^{2}-x^{3}\right) d u=0
\end{aligned}
$$

or $u^{3} d x+x\left(u^{2}-1\right) d u=0$. We separate the variables:

$$
\frac{\left(u^{2}-1\right) d u}{u^{3}}=-\frac{d x}{x}
$$

We integrate both sides:

$$
\int \frac{d u}{u}-\int u^{-3} d u=-\int \frac{d x}{x}, \ln |u|+\frac{1}{2 u^{2}}=-\ln |x|+\ln C
$$

we return to the function $w: \ln \left|\frac{w}{x}\right|+\frac{1}{2(w / x)^{2}}=\ln \frac{C}{|x|}$. We return to the function $y(x)=\frac{1}{w(x)}: \ln \left|\frac{1}{y x}\right|+\frac{y^{2} x^{2}}{2}=\ln \frac{C}{|x|}$. Note that in addition to this solution, the original equation has a particular solution $y=0$.

$$
\begin{aligned}
& \text { Ex.2.2.2. } \\
& \text { PROBLEMS FOR INDEPENDENT } \\
& \text { SOLVING }
\end{aligned}
$$

Integrate the following equations.