Example 5. Solve the equation $\left(3 x+2 y+y^{2}\right) d x+\left(x+4 x y+5 y^{2}\right) d y=0$, if its integrating factor has the form $\mu=\varphi\left(x+y^{2}\right)$.
Solution. Let $z=x+y^{2}$, then $\mu=\varphi(z)$, and, consequently,

$$
\frac{\partial \ln \mu}{\partial x}=\frac{d \ln \mu}{d z} \cdot \frac{\partial z}{\partial x}=\frac{\partial \ln \mu}{d z}, \quad \frac{\partial \ln \mu}{\partial y}=\frac{d \ln \mu}{d z} \cdot \frac{\partial z}{\partial y}=\frac{\partial \ln \mu}{d z} \cdot 2 y
$$

The equation (5) for finding the integrating factor will have the form

$$
(N-2 M y) \frac{d \ln \mu}{d z}=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}, \quad \text { or } \quad \frac{d \ln \mu}{d z}=\frac{(\partial M / \partial y)-(\partial N / \partial x)}{N-2 M y}
$$

Since $M=3 x+2 y+y^{2}, N=x+4 x y+5 y^{2}$, then

$$
\frac{(\partial M / \partial y)-(\partial N / \partial x)}{N-2 M y}=\frac{1}{x+y^{2}}=\frac{1}{z}
$$

and, therefore, $\frac{d \ln \mu}{d z}=\frac{1}{z}$, from which $\mu=z$, i.e., $\mu=x+y^{2}$. Multiplying the given equation by $\mu=x+y^{2}$, we get

$$
\left(3 x^{2}+2 x y+4 x y^{2}+2 y^{3}+y^{4}\right) d x+\left(x^{2}+4 x^{2} y+6 x y^{2}+4 x y^{3}+5 y^{4}\right) d y=0
$$

This is an equation in total differentials and its general integral according to (3) will be

$$
\begin{aligned}
& \int_{x_{0}}^{x}\left(3 x^{2}+2 x y+4 x y^{2}+2 y^{3}+y^{4}\right) d x+\int_{y_{0}}^{y}\left(x_{0}^{2}+4 x_{0}^{2} y+6 x_{0} y^{2}+4 x_{0} y^{3}+5 y^{4}\right) d y=C, \\
& \text { or } \\
& \quad x^{3}+x^{2} y+2 x^{2} y^{2}+2 x y^{3}+x y^{4}+y^{5}=\widehat{C}
\end{aligned}
$$

where

$$
\widehat{C}=C+x_{0}^{2} y_{0}+2 x_{0}^{2} y_{0}^{2}+2 x_{0} y_{0}^{3}+x_{0} y_{0}^{4}+y_{0}^{5}+x_{0}^{3}
$$

After some transformations, we will have

$$
(x+y)\left(x+y^{2}\right)^{2}=\widehat{C}
$$

## Problems for Independent Solution

Integrate the following equations:
189. $\left(1-x^{2} y\right) d x+x^{2}(y-x) d y=0$,

$$
\mu=\varphi(x)
$$

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

$\mu=\varphi(x)$.
191. $\left(x+y^{2}\right) d x-2 x y d y=0$,

$\mu=\varphi(x)$.
192. $\left(2 x^{2} y+2 y+5\right) d x+\left(2 x^{3}+2 x\right) d y=0$,

$\mu=\varphi(x)$.

193. $\left(x^{4} \ln x-2 x y^{3}\right) d x+3 x^{2} y^{2} d y=0$,
194. $(x+\sin x+\sin y) d x+\cos y d y=0$,
195. $\left(2 x y^{2}-3 y^{3}\right) d x+\left(7-3 x y^{2}\right) d y=0$,
196. $\left(3 y^{2}-x\right) d x+\left(2 y^{3}-6 x y\right) d y=0$,
197. $\left(x^{2}+y^{2}+1\right) d x-2 x y d y=0$,
198. $x d x+y d y+x(x d y-y d x)=0$, $\mu=\varphi(x)$.

$\mu=\varphi(x)$.

$\mu=\varphi(y)$.

$\mu=\varphi\left(x+y^{2}\right)$.

$\mu=\varphi\left(y^{2}-x^{2}\right)$.

$\mu=\varphi\left(x^{2}+y^{2}\right)$.

## § 8. First-Order Differential Equations Not Solved for the Derivative

$1^{\circ}$. First-order equations of $n$-th degree with respect to $\boldsymbol{y}^{\prime}$. Let us have a differential equation

$$
\left(y^{\prime}\right)^{n}+p_{1}(x, y)\left(y^{\prime}\right)^{n-1}+\ldots+p_{n-1}(x, y) y^{\prime}+p_{n}(x, y)=0
$$

Solve this equation with respect to $y^{\prime}$. Let

$$
y^{\prime}=f_{1}(x, y), \quad y^{\prime}=f_{2}(x, y), \quad \ldots, \quad y^{\prime}=f_{k}(x, y) \quad(k \leqslant n)
$$

- be the real solutions of equation (1).

The general integral of equation (1) will be expressed by the set of integrals:

$$
\Phi_{1}(x, y, C)=0, \quad \Phi_{2}(x, y, C)=0, \quad \ldots, \quad \Phi_{k}(x, y, C)=0
$$

where $\Phi_{i}(x, y, C)=0$ is the integral of the equation $y^{\prime}=f_{i}(x, y)(i=1, \ldots, k)$.

Thus, through each point of the domain where $y^{\prime}$ takes real values, there pass $k$ integral curves.