Solve the linear equations by considering y as a function of x, that is, \displaystyle{y}={y}{\left({x}\right)}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}-{2}\frac{x}{{{1}+{x}^{2}}}{y}={x}^{2}

Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
${\displaystyle y=y\left(x\right).\frac{dy}{dx}-\frac{2x}{1+x^2}y=x^2}$
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
${\displaystyle \int \frac{dy}{y}=\int \frac{2xdx}{1+x^2}}$
Let’s solve the integral on the right side.
${\displaystyle \int \frac{2xdx}{1+x^2}=|1+x^2=t\Rightarrow 2xdx=dt|}$
${\displaystyle =\int \frac{dt}{t}}$
${\displaystyle =\ln \left|t\right|+x}$
${\displaystyle =\ln |1+x^2|+c\wedge 1+x^2>0\mathrm{\forall }x\in \mathbb{R}}$
Therefore
${\displaystyle In\left|y\right|=\ln (1+x^2)+c}$.
By taking exponents, we obtain
${\displaystyle \ln \left|y\right|={\ln }^{1+x^2+c}=(1+x^2)\cdot e^c}$
Hence,we obtain
${\displaystyle y=C(1+x^2)}$
where ${\displaystyle C=\pm e^c\text{and}y,=1+x^2}$ is the complementary solution.
Next, we need to find the particular solution ${\displaystyle y_p}$.
Therefore, we consider ${\displaystyle u{y}_c}$, and try to find u, a function of x, that will make this work.
Let’s assume that ${\displaystyle u{y}_c}$ is a solution of the given equation. Hence, it satisfies the given equation.
Substituting ${\displaystyle u{y}_c}$ and its derivative in the equation gives
${\displaystyle \left(u^{\prime }{y}_c\right)-u{y}_c\frac{2x}{1+x^2}=x^2}$
${\displaystyle u^{\prime }{y}_c+u{y}_c^{\prime }\frac{2x}{1+x^2}=x^2}$
${\displaystyle u^{\prime }{y}_c+u\left(y_c^{\prime }\text{oprace}\frac{2x}{1+x^2}\right)=0\text{since}{y}_c\text{is a solution}=x^2}$
Therefore,
${\displaystyle u^{\prime }{y}_c=x^2\Rightarrow u^{\prime }=\frac{x^2}{y_c}}$
which gives
${\displaystyle u=\int \frac{x^2}{y_c}dx}$
Now, we can find the function u:
${\displaystyle u=\int \frac{x^2+1-1}{1+x^2}}$