4.59. $(2 y-3) d x+\left(2 x+3 y^{2}\right) d y=0, y(0)=1$.

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4.59. $(2 y-3) d x+\left(2 x+3 y^{2}\right) d y=0, y(0)=1$.
Solution. Step 1. Check condition (4.5):

$$
\begin{gathered}
P(x, y)=2 y-3, Q(x, y)=2 x+3 y^{2}, \\
\frac{\partial P}{\partial y}=\frac{\partial(2 y-3)}{\partial y}=2 ; \quad \frac{\partial Q}{\partial x}=\frac{\partial\left(2 x+3 y^{2}\right)}{\partial x}=2 ; \quad \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}=2
\end{gathered}
$$

- condition (4.5) is satisfied.

Step 2. Write down:

$$
\begin{gathered}
U(x, y)=\int P(x, y) d x+\varphi(y)= \\
=\int(2 y-3) d x+\varphi(y)=2 y \int d x-3 \int d x+\varphi(y)=2 y x-3 x+\varphi(y)
\end{gathered}
$$

Step 3. Since

$$
\frac{\partial U}{\partial y}=2 x+3 y^{2}=\frac{\partial(2 x y-3 x+\varphi(y))}{\partial y}=2 x+\frac{d \varphi}{d y}
$$

then $\frac{d \varphi(y)}{d y}=3 y^{2}$, hence

$$
\varphi(y)=\int 3 y^{2} d y+C=3 \frac{y^{3}}{3}+C=y^{3}+C .
$$

We obtained the general solution: $U(x, y)=2 x y-3 x+y^{2}+C=0$.

Solution of the Cauchy problem with the initial condition:

$$
U(x, y)=2 x y-3 x+y^{2}-1=0
$$

Step 1. Now first perform the integration: $Q(x, y)=2 x+3 y^{2}$, with respect to $y$.

Step 2. Write down:

$$
\begin{aligned}
& U(x, y)=\int Q(x, y) d y+\vartheta(x)=\int\left(2 x+3 y^{2}\right) d y+\vartheta(x)= \\
& \quad=2 x \int d y+3 \int y^{2} d y+9(x)=2 y x+y^{3}+\varphi(y)
\end{aligned}
$$

Step 3. Since

$$
\frac{\partial U}{\partial x}=2 y-3=\frac{\partial\left(2 x y+y^{3}+\vartheta(x)\right)}{\partial x}=2 y+\frac{d \vartheta}{d x},
$$

then $\frac{d 9(x)}{d x}=-3$, hence

$$
9(x)=-\int 3 d x+C_{1}=-3 x+C_{1}=-3 x+C_{1}
$$

We obtained the general solution: $U(x, y)=2 x y-3 x+y^{2}+C_{1}=0$.