Solve the linear equations by considering y as a function of x, that is, \displaystyle{y}={y}{\left({x}\right)}.{x}{y}'+{\left({1}+{x}\right)}{y}={e}^{{-{x}}} \sin{{2}}{x}

Carol Gates 2020-10-28 Answered

Solve the linear equations by considering y as a function of x, that is, \(\displaystyle{y}={y}{\left({x}\right)}.{x}{y}'+{\left({1}+{x}\right)}{y}={e}^{{-{x}}} \sin{{2}}{x}\)

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Expert Answer

Elberte
Answered 2020-10-29 Author has 27553 answers

Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
\(\displaystyle{x}{y}'=-{\left({1}+{x}\right)}{y}\Leftrightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=-\frac{{{\left({1}+{x}\right)}{y}}}{{x}}\Leftrightarrow\frac{d}{{y}}=-\frac{{{\left({1}+{x}\right)}{\left.{d}{x}\right.}}}{{x}}\)
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
\(\displaystyle{x}{y}'=-{\left({1}+{x}\right)}{y}\Leftrightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=-\frac{{{\left({1}+{x}\right)}{y}}}{{x}}\Leftrightarrow\frac{d}{{y}}=-\frac{{{\left({1}+{x}\right)}{\left.{d}{x}\right.}}}{{x}}\)
Let’s solve the integral on the right side.
\(\displaystyle\int\frac{{{\left({1}+{x}\right)}{\left.{d}{x}\right.}}}{{x}}=\int{\left(\frac{1}{{x}}+\frac{x}{{x}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\left(\frac{1}{{x}}+{1}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int\frac{{\left.{d}{x}\right.}}{{x}}+\int{\left.{d}{x}\right.}\)
\(\displaystyle={\ln}{\left|{x}\right|}+{x}+{c}\)
Therefore,
\(\displaystyle{\ln}{\left|{y}\right|}=-{\ln}{\left|{x}\right|}-{x}+{c}\)
By taking exponents, we obtain
\(\displaystyle{\left|{y}\right|}={e}^{{-{\ln}{\left|{x}\right|}-{x}+{c}}}=\frac{1}{{{\left|{x}\right|}{e}^{x}}}\cdot{e}\cdot{c}\)
Hence,we obtain
\(\displaystyle{y}=\frac{C}{{{x}{e}^{x}}}\)
where \(\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}=\frac{1}{{{x}{e}^{x}}}\) is the complementary solution.
Next, we need to find the particular solution \(\displaystyle{y}_{{p}}\).
Therefore, we consider \(\displaystyle{u}{y}_{{c}}\), and try to find u, a function of x, that will make this work.
Let’s assume that \(\displaystyle{u}{y}_{{c}}\), is a solution of the given equation. Hence, it satisfies the given equation.
Write the equation in the standard form (divide it by x 4 0)
\(\displaystyle{y}'+\frac{{{\left({1}+{x}\right)}{y}}}{{x}}=\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}\)
Substituting \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\) and its derivative in the equation gives
\(\displaystyle{u}{y}_{{c}}\)
\(\displaystyle{\left({u}{y}_{{c}}\right)}'+\frac{{{\left({1}+{x}\right)}{u}{y}_{{c}}}}{{x}}=\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}\)
\(\displaystyle{u}'{y}_{{c}}+{u}{\left({y}'{c}+\underbrace{{\frac{{{\left({1}+{x}\right)}{y}_{{c}}}}{{x}}}}\right)}{{={0}\ \text{since}\ {y}_{{c}}\text{is a solution}}}\)
Therefore,
\(\displaystyle{u}'{y}_{{c}}=\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}\Rightarrow{u}'=\frac{{\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}}}{{y}_{{c}}}\)
which gives
\(\displaystyle{u}=\int\frac{{\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}}}{{{y}_{{c}}}}{\left.{d}{x}\right.}\)
Now, we can find the function u :
\(\displaystyle=\int\frac{{{{x}}{{e}}^{x}\cdot{e}^{{-{x}}} \sin{{2}}{x}}}{{{{x}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=\int \sin{{2}}{x}{\left.{d}{x}\right.}\)
\(\displaystyle=-\frac{1}{{2}} \cos{{2}}{x}+{e}\)
Since we need to find only one function that will make this work, we don’t need to introduce the constant of integration c. Hence,
\(\displaystyle{u}=-\frac{1}{{2}} \cos{{2}}{x}\)
Recall that \(\displaystyle{y}_{{p}}={u}{y}_{{c}}\). Therefore,
\(\displaystyle{y}_{{p}}-\frac{1}{{2}}\cdot\frac{{ \cos{{2}}{x}}}{{{x}{e}^{x}}}\)
The general solution is
\(\displaystyle{y}={C}{y}_{{c}}+{y}_{{p}}\)
\(\displaystyle=\frac{C}{{{x}{e}^{x}}}-\frac{1}{{2}}\cdot\frac{{ \cos{{2}}{x}}}{{{x}{e}^{x}}}\)
\(\displaystyle=\frac{{{C}-\frac{1}{{2}}\cdot \cos{{2}}{x}}}{{{x}{e}^{x}}}\)
Integrating Factor technique
First, write the equation in the standard form (divide it by \(\displaystyle{x}\ne{0}\))
\(\displaystyle{y}'+\frac{{{\left({1}+{x}\right)}{y}}}{{x}}=\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}\)
This equation is linear with \(\displaystyle{P}{\left({x}\right)}=\frac{{{1}+{x}}}{{x}}{\quad\text{and}\quad}{Q}{\left({x}\right)}=\frac{{{e}^{{-{x}}} \sin{{2}}{x}}}{{x}}\)
, Hence,
\(\displaystyle{h}=\int{P}{\left.{d}{x}\right.}\int\frac{{{1}+{x}}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{x}\)
So, an integrating factor is
\(\displaystyle{e}^{h}={e}^{{{\ln}{\left|{x}\right|}+{x}}}={x}{e}^{x}\)
and the general solution is
\(\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle=\frac{1}{{{x}{e}^{x}}}{\left({c}+\int\frac{{{1}+{x}}}{{l{{x}}}}{{x}}{e}^{x}\right)}\)
\(\displaystyle=\frac{1}{{{x}{e}^{x}}}{\left({c}+\int\frac{{{{e}}^{{-{x}}} \sin{{2}}{x}}}{{{{x}}}}\cdot l{{x}}{{e}}^{x}\right)}\)
\(\displaystyle=\frac{{{c}-\frac{1}{{2}}\cdot \cos{{2}}{x}}}{{x}}{e}^{x}\)

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