Example 5. Integrate the equation

$$
\left(1+x^{2}\right) y^{\prime \prime}-2 x y^{\prime}=0
$$
Solution. This is an equation of the form (10.11), since the desired function $y$ does not explicitly appear in it. Let $y^{\prime}=p$, then

$$
y^{\prime \prime}=\frac{d y^{\prime}}{d x}=\frac{d p}{d x}
$$

and the equation takes the form

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

Separating the variables, we get

$$
\frac{d p}{p}=\frac{2 x d x}{1+x^{2}}
$$

Integrating, we find

$$
\ln |p|=\ln \left(1+x^{2}\right)+\ln \left|C_{1}\right|
$$

from which

$$
p=C_{1}\left(1+x^{2}\right) \text {. }
$$

Since

$$
p=\frac{d y}{d x}, \text { then } \frac{d y}{d x}=C_{1}\left(1+x^{2}\right)
$$

and, consequently, the general solution is given by the formula

$$
y=C_{1}\left(\frac{x^{3}}{3}+x\right)+C_{2}
$$

Remark. When separating the variables, we assumed that $p \neq 0$, $1+x^{2} \neq 0$, so we could have lost the solutions $p=0,1+x^{2}=0$. The first equality implies that $y^{\prime}=0$ and $y=C$. The function $y=C$ is a solution to the original equation, which can be verified directly. These solutions are obtained from the general solution when $C_{1}=0$. The second equality is impossible for real $x$, it does not define a function that is a solution to the given equation.