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}

chillywilly12a 2021-02-24 Answered

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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Expert Answer

lobeflepnoumni
Answered 2021-02-25 Author has 16396 answers
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 don’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 \(\displaystyle{y}=\frac{c}{{x}}+\frac{{{x}^{2}}}{{3}}\)
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