# Solve the linear equations by considering y as a function of x, that is, y = y(x). displaystylefrac{{{left.{d}{y}right.}}}{{{left.{d}{x}right.}}}+{left(frac{1}{{x}}right)}{y}={x}

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

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Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
$$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-\frac{y}{{x}}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}=-\frac{{{\left.{d}{x}\right.}}}{{x}}$$
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
$$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=-\int\frac{{{\left.{d}{x}\right.}}}{{x}}$$
Which is
$$\displaystyle \ln{{\left|{{y}}\right|}}=- \ln{{\left|{{x}}\right|}}+{c}$$
By taking exponents, we obtain
$$\displaystyle{\left|{{y}}\right|}={e}^{{- \ln{{\left|{{x}}\right|}}+{c}}}=\frac{1}{{{\left|{{x}}\right|}}}\cdot{e}^{c}$$
Hence,we obtain
$$\displaystyle{y}=\frac{1}{{C}}{x}$$
where $$\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}=\frac{1}{{x}}$$ is the complementary solution .
Next, we need to find the particular solution $$y_p$$.
Therefore, we consider $$uy_c$$, and try to find u, a function of x, that will make this work.
Let's assume that $$uy_c$$ is a solution of the given equation. Hence, it satisfies the given equation.
Substituting $$uy_c$$, and its derivative in the equation gives
$$\displaystyle{\left({u}{y}_{{c}}\right)}'+\frac{{{u}{y}_{{c}}}}{{x}}={x}$$
$$\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}+\frac{{{u}{y}_{{c}}}}{{x}}={x}$$
$$\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}+\frac{{{y}_{{c}}}}{{x}}\right)}}}_{{\text{=0 }\ }}={x}$$
since $$y_c$$ is a solution
Therefore,
$$\displaystyle{u}'{y}_{{c}}={x}\Rightarrow{u}'=\frac{x}{{{y}_{{c}}}}$$
which gives
$$\displaystyle{u}=\int\frac{x}{{{y}_{{c}}}}{\left.{d}{x}\right.}$$
Now, we can find the function u:
$$\displaystyle{u}=\int{\left(\frac{x}{{\frac{1}{{x}}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{x}^{2}{\left.{d}{x}\right.}$$
$$\displaystyle=\frac{{{x}^{3}}}{{3}}+{c}$$
Since we need to find only one function that will male this work, we donâ€™t need to introduce the constant of integration c. Hence,
$$\displaystyle{u}=\frac{{{x}^{3}}}{{3}}$$
Recall that $$y_p=uy_c$$
Therefore
$$\displaystyle{y}_{{p}}=\frac{1}{{x}}\cdot\frac{{{x}^{3}}}{{3}}$$
$$\displaystyle=\frac{{{x}^{2}}}{{3}}$$
The general solution is
$$\displaystyle{y}={C}{y}_{{c}}+{y}_{{p}}$$
$$\displaystyle=\frac{C}{{x}}+\frac{{{x}^{2}}}{{3}}$$
Integrating Factor technique
This equation is linear with $$\displaystyle{P}{\left({x}\right)}=\frac{1}{{x}}{\quad\text{and}\quad}{Q}{\left({x}\right)}={x}$$
Hence, $$\displaystyle{h}=\int\frac{1}{{x}}= \ln{{\left|{{x}}\right|}}$$
So, an integrating factor is $$\displaystyle{e}^{h}={e}^{{ \ln{{\left|{{x}}\right|}}}}={x}$$
and the general solution is
$$\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle=\frac{1}{{x}}{\left({c}+\int{x}\cdot{x}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle=\frac{1}{{x}}{\left({c}+\int{x}^{2}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle=\frac{1}{{x}}{\left({c}+\frac{{{x}^{3}}}{{3}}\right)}$$
$$\displaystyle=\frac{c}{{x}}+\frac{{{x}^{2}}}{{3}}$$
Answer $$\displaystyle{y}=\frac{c}{{x}}+\frac{{{x}^{2}}}{{3}}$$