# 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}

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

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Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
$y=y\left(x\right).\frac{dy}{dx}-\frac{2x}{1+{x}^{2}}y={x}^{2}$
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
$\int \frac{dy}{y}=\int \frac{2xdx}{1+{x}^{2}}$
Let’s solve the integral on the right side.
$\int \frac{2xdx}{1+{x}^{2}}=|1+{x}^{2}=t⇒2xdx=dt|$
$=\int \frac{dt}{t}$
$=\mathrm{ln}|t|+x$
$=\mathrm{ln}|1+{x}^{2}|+c\wedge 1+{x}^{2}>0\mathrm{\forall }x\in \mathbb{R}$
Therefore
$In|y|=\mathrm{ln}\left(1+{x}^{2}\right)+c$.
By taking exponents, we obtain
$\mathrm{ln}|y|={\mathrm{ln}}^{1+{x}^{2}+c}=\left(1+{x}^{2}\right)\cdot {e}^{c}$
Hence,we obtain
$y=C\left(1+{x}^{2}\right)$
where $C\phantom{\rule{0.222em}{0ex}}=±{e}^{c}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y,=1+{x}^{2}$ is the complementary solution.
Next, we need to find the particular solution ${y}_{p}$.
Therefore, we consider $u{y}_{c}$, and try to find u, a function of x, that will make this work.
Let’s assume that $u{y}_{c}$ is a solution of the given equation. Hence, it satisfies the given equation.
Substituting $u{y}_{c}$ and its derivative in the equation gives
$\left({u}^{\prime }{y}_{c}\right)-u{y}_{c}\frac{2x}{1+{x}^{2}}={x}^{2}$
${u}^{\prime }{y}_{c}+u{y}_{c}^{\prime }\frac{2x}{1+{x}^{2}}={x}^{2}$

Therefore,
${u}^{\prime }{y}_{c}={x}^{2}⇒{u}^{\prime }=\frac{{x}^{2}}{{y}_{c}}$
which gives
$u=\int \frac{{x}^{2}}{{y}_{c}}dx$
Now, we can find the function u:
$u=\int \frac{{x}^{2}+1-1}{1+{x}^{2}}$