4.61. Integrate the equation

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

by finding the integrating factor.
Solution.

$$
P(x, y)=x^{2} y^{2}-1, Q(x, y)=2 x^{3} y, \frac{\partial P}{\partial y}=2 x^{2} y, \frac{\partial Q}{\partial x}=6 x^{2} y
$$

Thus, the equality $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ does not hold, and the equation is not an exact differential equation. First, we will try to find an integrating factor that depends on $y$. For this, we form the equation:

$$
\begin{gathered}
\frac{1}{P(x, y)}\left(\frac{\partial P(x, y)}{\partial y}-\frac{\partial Q(x, y)}{\partial x}\right)= \\
=\frac{1}{x^{2} y^{2}-1} \cdot\left(2 x^{2} y-6 x^{2} y\right)=-\frac{4 x^{2} y}{x^{2} y^{2}-1} .
\end{gathered}
$$

We did not obtain an expression that depends only on $y$. We will now look for an integrating factor that depends on $x$. We form the equation:

$$
\begin{gathered}
\frac{1}{Q(x, y)}\left(\frac{\partial P(x, y)}{\partial y}-\frac{\partial Q(x, y)}{\partial x}\right)= \\
=\frac{1}{2 x^{3} y} \cdot\left(2 x^{2} y-6 x^{2} y\right)=\frac{-4 x^{2} y}{2 x^{3} y}=-\frac{2}{x} .
\end{gathered}
$$

Since this expression depends only on $x$, an integrating factor $\mu(x)$ exists and is determined from the equation

$$
\frac{d \mu}{\mu}=-\frac{2}{x} \cdot d x
$$

We have:

$$
\int \frac{d \mu}{\mu}=-2 \int \frac{d x}{x} ; \ln |\mu|=-2 \ln |x|+\ln C, \mu=\frac{C}{x^{2}}
$$

We choose a particular non-zero solution from this general solution by setting, for example, $C=1: \mu=\frac{1}{x^{2}}$.

Now, multiply both sides of the original equation by the integrating factor:

$$
\frac{1}{x^{2}} \cdot\left(x^{2} y^{2}-1\right) d x+\frac{1}{x^{2}} \cdot 2 x^{3} y d y=\frac{1}{x^{2}} \cdot 0
$$

We get:

$$
\left(y^{2}-\frac{1}{x^{2}}\right) d x+2 x y d y=0
$$

## We have:

$$
P(x, y)=y^{2}-\frac{1}{x^{2}}, \quad Q(x, y)=2 x y, \frac{\partial P}{\partial y}=2 y, \quad \frac{\partial Q}{\partial x}=2 y
$$

Thus, the equality $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ holds, and the equation is an exact differential equation. We first seek a representation for $U(x, y)$ based on the fact that

$$
d U=\frac{\partial P}{\partial x} d x+\frac{\partial Q}{\partial y} d y \text { and } \frac{\partial U}{\partial y}=Q(x, y)=2 x y
$$

from which

$$
\begin{gathered}
U(x, y)=\int 2 x y d y+\varphi(x) ; U(x, y)= \\
=x y^{2}+\varphi(x) ; \quad \frac{\partial U}{\partial x}=y^{2}+\frac{d \varphi}{d x}=P(x, y)=y^{2}-\frac{1}{x^{2}}
\end{gathered}
$$

For $\varphi(x)$, we obtain a separable differential equation:

$$
\frac{d \varphi}{d x}=-\frac{1}{x^{2}} ; \quad d \varphi=-\frac{d x}{x^{2}} ; \int d \varphi=-\int \frac{d x}{x^{2}} ; \varphi=\frac{1}{x^{2}}-C .
$$

The general solution of the original equation is:

$$
U(x, y)=x y^{2}+\frac{1}{x}-C=0 \text { or } x y^{2}+\frac{1}{x}=C
$$

### 4.2.4.2.

PROBLEMS FOR INDEPENDENT SOLUTION WITH ANSWERS

Integrate the differential equations, first ensuring that they are exact differential equations.