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}

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}

Question
Solve the linear equations by considering y as a function of x, that is, \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}.{\left\lbrace{x}\right\rbrace}{\left\lbrace{y}\right\rbrace}'+{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\)

Answers (1)

2020-10-29
Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}{\left\lbrace{y}\right\rbrace}'=-{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}\Leftrightarrow{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}}=-{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\Leftrightarrow{\frac{{{d}}}{{{\left\lbrace{y}\right\rbrace}}}}=-{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\)
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
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\int{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace{y}\right\rbrace}}}}-\int{\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}},\)
Let’s solve the integral on the right side.
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\int{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}=\int{\left\lbrace{\left({\frac{{{1}}}{{{\left\lbrace{x}\right\rbrace}}}}+{\frac{{{x}}}{{{\left\lbrace{x}\right\rbrace}}}}\right)}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\left\lbrace{\left({\frac{{{1}}}{{{\left\lbrace{x}\right\rbrace}}}}+{\left\lbrace{1}\right\rbrace}\right)}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}+\int{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}+{\left\lbrace{x}\right\rbrace}+{\left\lbrace{c}\right\rbrace}\)
Therefore,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}=-{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}-{\left\lbrace{x}\right\rbrace}+{\left\lbrace{c}\right\rbrace}.\)
By taking exponents, we obtain
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace\le{f}{t}{\left|{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}-{\left\lbrace{x}\right\rbrace}+{\left\lbrace{c}\right\rbrace}\right\rbrace}}}={\frac{{{1}}}{{{\left\lbrace{\left\lbrace\le{f}{t}{\left|{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\cdot{\left\lbrace{e}\right\rbrace}\cdot{\left\lbrace{c}\right\rbrace}\)
Hence,we obtain
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\frac{{{C}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\)
where \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{C}\right\rbrace}\:=\pm{\left\lbrace{e}\right\rbrace}^{{{c}}}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}={\frac{{{1}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\) is the complementary solution.
Next, we need to find the particular solution \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{p}\right\rbrace}}}\).
Therefore, we consider \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\), and try to find u, a function of x, that will make this work.
Let’s assume that \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\), 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)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}'+{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\)
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
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right)}\right\rbrace}'+{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}+{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}'_{{{\left\lbrace{c}\right\rbrace}}}+{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}+{\left\lbrace{u}\right\rbrace}{\left\lbrace{\left({\left\lbrace{y}\right\rbrace}'{\left\lbrace{c}\right\rbrace}+\underbrace{{{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\right\rbrace}}}\right)}\right\rbrace}_{{{\left\lbrace={\left\lbrace{0}\right\rbrace}\ \text{since}\ {\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\text{is a solution}\right\rbrace}}}\)
Therefore,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}={\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\Rightarrow{\left\lbrace{u}\right\rbrace}'={\frac{{{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\right\rbrace}}}{{{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}}}}\)
which gives
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}=\int{\frac{{{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
Now, we can find the function u :
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}=\int{\frac{{{\left\lbrace{\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\right\rbrace}}}{{{\left\lbrace{\frac{{{1}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\frac{{{\left\lbrace\cancel{{{\left\lbrace{x}\right\rbrace}}}\cancel{{{\left\lbrace{e}\right\rbrace}}}^{{{x}}}\cdot{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace\cancel{{{\left\lbrace{x}\right\rbrace}}}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=-{\frac{{{1}}}{{{\left\lbrace{2}\right\rbrace}}}}{\cos{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{e}\right\rbrace}\)
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,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}=-{\frac{{{1}}}{{{\left\lbrace{2}\right\rbrace}}}}{\cos{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\)
Recall that \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{p}\right\rbrace}}}={\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\). Therefore,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{p}\right\rbrace}}}-{\frac{{{1}}}{{{\left\lbrace{2}\right\rbrace}}}}\cdot{\frac{{{\left\lbrace{\cos{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\)
The general solution is
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{C}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}+{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{p}\right\rbrace}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\frac{{{C}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}-{\frac{{{1}}}{{{\left\lbrace{2}\right\rbrace}}}}\cdot{\frac{{{\left\lbrace{\cos{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\frac{{{\left\lbrace{\left\lbrace{C}\right\rbrace}-{\frac{{{1}}}{{{\left\lbrace{2}\right\rbrace}}}}\cdot{\cos{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}\)
Integrating Factor technique
First, write the equation in the standard form (divide it by \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}\ne{\left\lbrace{0}\right\rbrace}\))
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}'+{\frac{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{y}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\)
This equation is linear with \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{P}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}={\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{Q}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}={\frac{{{\left\lbrace{\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}\)
, Hence,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{h}\right\rbrace}=\int{\left\lbrace{P}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\int{\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}={\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\)
So, an integrating factor is
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{e}\right\rbrace}^{{{h}}}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\)
and the general solution is
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{h}\right\rbrace}\right\rbrace}}}{\left\lbrace{\left({\left\lbrace{c}\right\rbrace}+\int{\left\lbrace{Q}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{h}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right)}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\frac{{{1}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}{\left\lbrace{\left({\left\lbrace{c}\right\rbrace}+\int{\frac{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace\cancel{{{\left\lbrace{x}\right\rbrace}}}\right\rbrace}}}}\cancel{{{\left\lbrace{x}\right\rbrace}}}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right)}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\frac{{{1}}}{{{\left\lbrace{\left\lbrace{x}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{x}}}\right\rbrace}}}}{\left\lbrace{\left({\left\lbrace{c}\right\rbrace}+\int{\frac{{{\left\lbrace\cancel{{{\left\lbrace{e}\right\rbrace}}}^{{{\left\lbrace-{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{\sin{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace\cancel{{{\left\lbrace{x}\right\rbrace}}}\right\rbrace}}}}\cdot\cancel{{{\left\lbrace{x}\right\rbrace}}}\cancel{{{\left\lbrace{e}\right\rbrace}}}^{{{x}}}\right)}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\frac{{{\left\lbrace{\left\lbrace{c}\right\rbrace}-{\frac{{{1}}}{{{\left\lbrace{2}\right\rbrace}}}}\cdot{\cos{{\left\lbrace{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{x}\right\rbrace}}}}{\left\lbrace{e}\right\rbrace}^{{{x}}}\)
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