899. Check that the given expression is a total differential of the function $u(x, y)$, and find $u$:
1) $\left.\left(2 x-3 y^{2}+1\right) d x+(2-6 x y) d y ; \quad 2\right)\left(e^{x y}+5\right)(x d y+y d x)$;
2) $(1-\sin 2 x) d y-(3+2 y \cos 2 x) d x$.
Solution. 1) Let's denote the coefficients at differentials as $P=2 x-3 y^{2}+1, Q=2-6 x y$ and find $P_{y}^{\prime}=-6 y$ and $Q_{x}^{\prime}=-6 y$. Since here $P_{y}^{\prime}=Q_{x}^{\prime}$ and $P, Q, P_{y}^{\prime}, Q_{x}^{\prime}$ are continuous, the given expression is a total differential of some function $u$.

We find this function using formula (1), choosing point $A$ at the origin $O(0,0)$

$$
u=\int_{=1}^{x}(2 x+1) d x+\int_{"}^{u}(2-6 x y) d y+C=x^{2}+x+2 y-3 x y^{2}+C
$$

2) Transform the given differential expression to the form $P d x+Q d y$ and find $P_{u \prime}^{\prime}$ and $Q_{x}^{\prime}$:

$$
y\left(e^{x y}+5\right) d x+x\left(e^{x y}+5\right) d y ; \quad P_{y}^{\prime}=5+e^{x y}(1+x y)=Q_{x}^{\prime}
$$

The condition $P_{u}^{\prime}=Q_{x}^{\prime}$ is satisfied. The given expression is a total differential of some function $u(x, y)$.

We find this function using formula (2):

$$
\begin{gathered}
u=\int_{x_{0}}^{x} y\left(e^{x_{y}}+5\right) d x+\int_{y_{0}}^{y} x_{0}\left(e^{x_{0} y}+5\right) d y+C=e^{x y}+\left.5 x y\right|_{x_{u}} ^{x}+ \\
+-e^{x_{0} y}+\left.5 x_{0} y\right|_{y_{0}} ^{y}+C=e^{x y}+5 x y-e^{x_{0} y_{0}}-5 x_{0} y_{0}+C=e^{x^{y} y}+5 x y+C_{1} \\
\text { where } C_{1}=C-e^{x_{0} y_{0}}-5 x_{0} y_{0} \\
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\end{gathered}
$$

3) First, we find the partial derivatives

$$
\begin{gathered}
P_{y}^{\prime}=-(3+2 y \cos 2 x)_{y}^{\prime}=-2 \cos 2 x, \\
Q_{x}^{\prime}=(1-\sin 2 x)_{x}^{\prime}=-2 \cos 2 x
\end{gathered}
$$

and verify that they are identically equal and that the given expression is a total differential of some function $u(x, y)$. Then we find this function using the second method, integrating each partial differential $P d x$ and $Q d y$ separately.

a) $u=-\int(3+2 y \cos 2 x) d x=-3 x-y \sin 2 x+\varphi(y), \quad$ considering $y$ as a constant;

b) $u=\int(1-\sin 2 x) d y=y-y \sin 2 x+\psi(x)$, considering $x$ as a constant.

Combining these two expressions - adding to the known terms of the first expression the missing term, which depends only on $y$, from the second expression, we obtain one of the antiderivative functions, and by adding an arbitrary constant $C$ to it, we obtain the general expression of the antiderivative function for the given total differential $u=y-3 x-y \sin 2 x+C$.

In problems $900-905$, verify that the given differential expression is a total differential of some function $u(x, y)$ and then find $u$: