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

Question
Solve the linear equations by considering y as a function of x, that is, \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}.{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}}-{\left\lbrace{2}\right\rbrace}{\frac{{{x}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}{\left\lbrace{y}\right\rbrace}={\left\lbrace{x}\right\rbrace}^{{{2}}}\)

Answers (1)

2020-11-21
Variation of parameters
First, solve the linear homogeneous equation by separating variables.
Rearranging terms in the equation gives
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}}={\left\lbrace{2}\right\rbrace}{\frac{{{x}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}{\left\lbrace{y}\right\rbrace}\Leftrightarrow{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace{y}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\)
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
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\int{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace{y}\right\rbrace}}}}=\int{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\)
Let’s solve the integral on the right side.
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\int{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}={\left\lbrace\le{f}{t}{\left|{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}={\left\lbrace{t}\right\rbrace}\Rightarrow{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}={\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{t}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{t}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace{t}\right\rbrace}}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{t}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}+{\left\lbrace{x}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}{r}{i}{g}{h}{t}\right|}\right\rbrace}+{\left\lbrace{c}\right\rbrace}\wedge{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}{>}{\left\lbrace{0}\right\rbrace}\forall{\left\lbrace{x}\right\rbrace}\in{\mathbb{{{R}}}}\)
Therefore
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{I}\right\rbrace}{\left\lbrace{n}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}={\ln{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\right\rbrace}}}+{\left\lbrace{c}\right\rbrace}\).
By taking exponents, we obtain
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{y}\right\rbrace}{r}{i}{g}{h}{t}\right|}\right\rbrace}={\left\lbrace{\left\lbrace{\ln}\right\rbrace}^{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{c}\right\rbrace}\right\rbrace}}}=\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\cdot{\left\lbrace{e}\right\rbrace}^{{{c}}}\)
Hence,we obtain
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{C}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\)
where \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{C}\right\rbrace}\:=\pm{\left\lbrace{e}\right\rbrace}^{{{c}}}{\left\lbrace\quad\text{and}\quad\right\rbrace}{\left\lbrace{y}\right\rbrace},={\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\) 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.
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
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{\left({\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right)}\right\rbrace}-{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}={\left\lbrace{x}\right\rbrace}^{{{2}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}+{\left\lbrace{u}\right\rbrace}{\left\lbrace{y}\right\rbrace}'_{{{\left\lbrace{c}\right\rbrace}}}{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}={\left\lbrace{x}\right\rbrace}^{{{2}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}+{\left\lbrace{u}\right\rbrace}{\left\lbrace{\left({\left\lbrace{y}\right\rbrace}'_{{{\left\lbrace{c}\right\rbrace}}}\text{oprace}{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\right)}\right\rbrace}_{{{\left\lbrace={\left\lbrace{0}\right\rbrace}\ \text{since}\ {\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\text{is a solution}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}^{{{2}}}\)
Therefore,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}'{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}={\left\lbrace{x}\right\rbrace}^{{{2}}}\Rightarrow{\left\lbrace{u}\right\rbrace}'={\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}{{{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}}}}\)
which gives
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}=\int{\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
Now, we can find the function u:
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}=\int{\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{1}\right\rbrace}-{\left\lbrace{1}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}-{\frac{{{1}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\right)}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le=\int{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}-\int{\frac{{{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{x}\right\rbrace}-{\arctan{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}+{\left\lbrace{c}\right\rbrace}\)
Since we need to find only one function that will make this work, we dont need to introduce the constant of integration c. Hence,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{u}\right\rbrace}={\left\lbrace{x}\right\rbrace}-{\arctan{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}\)
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,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{p}\right\rbrace}}}={\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}-{\arctan{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}\right)}\right\rbrace}\)
The general solution is
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{C}\right\rbrace}{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{c}\right\rbrace}}}+{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{p}\right\rbrace}}}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}-{\arctan{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}\right)}\right\rbrace}+{\left\lbrace{C}\right\rbrace}{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\)
Integrationg Factor technique
This equation is linear with \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{P}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}=-{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}=-{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}{r}{i}{g}{h}{t}\right|}\right\rbrace}=-{\ln{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\right\rbrace}}}\)
Hence,
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{h}\right\rbrace}=\int{\left\lbrace{P}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}-\int{\frac{{{\left\lbrace{\left\lbrace{2}\right\rbrace}{\left\lbrace{x}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}=-{\left\lbrace{\ln}\right\rbrace}{\left\lbrace\le{f}{t}{\left|{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}{r}{i}{g}{h}{t}\right|}\right\rbrace}=-{\ln{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\right\rbrace}}}\)
So, an integrating factor is
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{e}\right\rbrace}^{{{h}}}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\ln{{\left\lbrace{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}\right\rbrace}}}\right\rbrace}}}={\frac{{{1}}}{{{\left\lbrace{\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right\rbrace}}}}\)
and the general solution is
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}\right)}\right\rbrace}={\left\lbrace{e}\right\rbrace}^{{{\left\lbrace-{\left\lbrace{h}\right\rbrace}\right\rbrace}}}{\left\lbrace{\left({\left\lbrace{c}\right\rbrace}+\int{\left\lbrace{Q}\right\rbrace}{\left\lbrace{e}\right\rbrace}^{{{h}}}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right)}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}{\left\lbrace{\left({\left\lbrace{c}\right\rbrace}+\int{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}.{\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}{\left\lbrace\le{f}{t}.{\left\lbrace{d}\right\rbrace}{\left\lbrace{x}\right\rbrace}{r}{i}{g}{h}{t}.\right\rbrace}\right)}\right\rbrace}\)
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le={\left\lbrace{\left({\left\lbrace{1}\right\rbrace}+{\left\lbrace{x}\right\rbrace}^{{{2}}}\right)}\right\rbrace}{\left\lbrace{\left({\left\lbrace{c}\right\rbrace}+{\left\lbrace{x}\right\rbrace}-{\arctan{{\left\lbrace{\left\lbrace{x}\right\rbrace}\right\rbrace}}}\right)}\right\rbrace}\)
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