# Solve the linear equations by considering y as a function of x, that is, y = y(x) 3xy'+y=12x Question
Second order linear equations Solve the linear equations by considering y as a function of x, that is,
$$y = y(x)$$
$$3xy'+y=12x$$ 2021-02-13
Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
$$\displaystyle{3}{x}{y}'=-{y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-\frac{y}{{{3}{x}}}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}=-\frac{{{\left.{d}{x}\right.}}}{{{3}{x}}}$$
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
$$\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=-\frac{1}{{3}}\int\frac{{{\left.{d}{x}\right.}}}{{x}}$$
Which is
$$\displaystyle \ln{{\left|{{y}}\right|}}=-\frac{1}{{3}} \ln{{\left|{{x}}\right|}}+{c}$$
By taking exponents, we obtain
$$\displaystyle{\left|{{y}}\right|}={e}^{{-\frac{1}{{3}} \ln{{\left|{{x}}\right|}}+{c}}}={\left|{{x}}\right|}^{{-\frac{1}{{3}}}}\cdot{e}^{c}$$
Hence,we obtain
$$\displaystyle{y}={C}{x}^{{-\frac{1}{{3}}}}$$
where $$\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={x}^{{-\frac{1}{{3}}}}$$ 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.
Write the equation in the standard form $$\displaystyle{\left(\text{divide iy by }\ {3}{x}\ne{0}\right)}$$
$$\displaystyle{y}'+\frac{y}{{{3}{x}}}={12}$$
Substituting $$uy_c$$, and its derivative in the equation gives
$$\displaystyle{\left({u}{y}_{{c}}\right)}'+\frac{{{u}{y}_{{c}}}}{{{3}{x}}}={4}$$
$$\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}+\frac{{{u}{y}_{{c}}}}{{{3}{x}}}={4}$$
$$\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}+\frac{{{y}_{{c}}}}{{{3}{x}}}\right)}}}_{{\text{=0 }\ }}={4}$$
since $$y_c$$ is a solution
Therefore,
$$\displaystyle{u}'{y}_{{c}}={4}\Rightarrow{u}'=\frac{{{4}}}{{{y}_{{c}}}}$$
which gives
$$\displaystyle{u}=\int\frac{{{4}}}{{{y}_{{c}}}}{\left.{d}{x}\right.}$$
Now, we can find the function u:
$$\displaystyle{u}=\int\frac{{{4}}}{{{x}^{{-\frac{1}{{3}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={4}\int{x}^{{-\frac{1}{{3}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={4}\cdot\frac{{{x}^{{\frac{1}{{3}}+{1}}}}}{{\frac{1}{{3}}+{1}}}+{c}$$
$$\displaystyle={4}\cdot\frac{{{x}^{{\frac{4}{{3}}}}}}{{\frac{4}{{3}}}}+{c}$$
$$\displaystyle={4}\cdot\frac{3}{{4}}{x}^{{\frac{4}{{3}}}}+{c}$$
$$\displaystyle={3}{x}^{{\frac{4}{{3}}}}+{c}$$
Since we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,
$$\displaystyle{u}={3}{x}^{{\frac{4}{{3}}}}$$
Recall that $$y_p=uy_c$$
Therefore
$$\displaystyle{y}_{{p}}={3}{x}^{{\frac{4}{{3}}}}\cdot{x}^{{-\frac{1}{{3}}}}$$
$$=9x$$
The general solution is
$$y=Cy_c+y_p$$
$$\displaystyle={C}{x}^{{-\frac{1}{{3}}}}+{3}{x}$$
Integration Factor technique
First , write the equation in the standard form (divide iy by 3x != 0)\)
$$\displaystyle{y}'+\frac{y}{{{3}{x}}}={4}$$
This equation is linear with $$\displaystyle{P}{\left({x}\right)}=\frac{1}{{{3}{x}}}{\quad\text{and}\quad}{Q}{\left({x}\right)}={4}$$
Hence,
$$\displaystyle{h}=\frac{1}{{3}}\int\frac{1}{{x}}{\left.{d}{x}\right.}=\frac{1}{{3}} \ln{{\left|{{x}}\right|}}$$
So, an integrating factor is
$$\displaystyle{e}^{h}={e}^{{\frac{1}{{3}} \ln{{\left|{{x}}\right|}}}}={x}^{{\frac{1}{{3}}}}$$
and the general solution is
$$\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle={x}^{{-\frac{1}{{3}}}}{\left({c}+\int{4}\cdot{x}^{{\frac{1}{{3}}}}{\left.{d}{x}\right.}\right)}$$
$$\displaystyle={x}^{{\frac{1}{{3}}}}{\left({c}+{3}{x}^{{\frac{4}{{3}}}}\right)}$$
$$\displaystyle={c}{x}^{{\frac{1}{{3}}}}+{3}{x}^{{\frac{4}{{3}}}}$$
Answer $$\displaystyle{y}={C}{x}^{{\frac{1}{{3}}}}+{3}{x}^{{\frac{4}{{3}}}}$$

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