1091. Solve the equations:
1) $(2 y-3) d x+\left(2 x+3 y^{2}\right) d y=0$
2) $(x+\ln |y|) d x+\left(1+\frac{x}{y}+\sin y\right) d y=0$.
Solution. 1) First, we ensure that the given equation is an equation in total differentials:

$$
P_{y}^{\prime}=(2 y-3)_{y}^{\prime}=2 ; \quad Q_{x}^{\prime}=\left(2 x+3 y^{2}\right)_{x}^{\prime}=2 ; \quad P_{y}^{\prime}=Q_{x}^{\prime}
$$

Then we find the indefinite integrals:

$\int P d x=\int(2 y-3) d x=2 x y-3 x+\varphi(y)$, treating $y$ as a constant, $\int Q d y=\int\left(2 x+3 y^{2}\right) d y=2 x y+y^{3}+\psi(x)$, treating $x$ as a constant.

Taking all known terms from the first result and adding the missing terms, which depend only on $y$, from the second result, we obtain the function $u(x, y)=2 x y-3 x+y^{3}$, the total differential of which is the left-hand side of the given differential equation. By equating it to an arbitrary constant, we obtain the general integral of the given equation:

$$
2 x y-3 x+y^{3}=C
$$[^44]

2) After verifying that the left-hand side of the given equation is the total differential of some function $u(x, y)$:

$$
(x+\ln |y|)_{y}^{\prime}=\frac{1}{y}==\left(1+\frac{x}{y}+\sin y\right)_{x}^{\prime},
$$

we then find this function by integrating each of its partial differentials separately:

$$
\begin{gathered}
u=\int(x+\ln |y|) d x=\frac{x^{2}}{2}+x \ln |y|+\varphi(y) \\
u=\int\left(1+\frac{x}{y}+\sin y\right) d y=y+x \ln |y|-\cos y+\psi(x)
\end{gathered}
$$

Next, we form the final expression for the function $u$ (we add the missing terms, which depend only on $y$, from the second expression to the known terms of the first expression) and, by equating it to an arbitrary constant $C$, we find the general integral of the given equation:

$$
\frac{x^{2}}{2}+x \ln |y|+y-\cos y=C
$$

Check that the following first-order equations are equations in total differentials and solve them.