4.60. $\left(3 x^{2} y^{2}+7\right) d x+2 x^{3} y d y=0, y(0)=1$.
Solution. $1-\check{i}$ step. Check condition (4.5):

$$
P(x, y)=3 x^{2} y^{2}+7, Q(x, y)=2 x^{3} y
$$

$2-\bar{i}$ step. Use formula (4.6):

$$
\begin{aligned}
& U(x, y)=\int_{x_{0}}^{x}\left(3 x^{2} y^{2}+7\right) d x+\int_{y_{0}}^{y} 2 x_{0}^{3} y d y=C, \\
& U(x, y)=y^{2} \int_{x_{0}}^{x}\left(3 x^{2} d x+7 \int_{x_{0}}^{x} d x+2 x_{0}^{3} \int_{y_{0}}^{y} y d y=C,\right. \\
& U(x, y)=\left.y^{2} x^{3}\right|_{x_{0}} ^{x}+\left.7 x\right|_{x_{0}} ^{x}+\left.x_{0}^{3} y^{2}\right|_{y_{0}} ^{y}=C, \\
& U(x, y)=y^{2}\left(x^{3}-x_{0}^{3}\right)+7\left(x-x_{0}\right)+x_{0}^{3}\left(y^{2}-y_{0}^{2}\right)=C, \\
& U(x, y)=y^{2} x^{3}+7 x=C+7 x_{0}+x_{0}^{3} y_{0}^{2} \equiv C_{1} .
\end{aligned}
$$

Here $x_{0}, y_{0}$ are a priori chosen numbers (constants). Introduce the notation $C+7 x_{0}+x^{3}{ }_{0} y^{2}{ }_{0} \equiv C_{1}$, then the final form of the general solution in implicit form will be $y^{2} x^{3}+7 x=C_{1}$.