Solve the linear equations by considering y as a function of x, that is, y = y(x) 3xy'+y=12x

Question
Solve the linear equations by considering y as a function of x, that is,
\(y = y(x)\)
\(3xy'+y=12x\)

Answers (1)

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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