Answer

Rearranging the equation, we have

\[ \begin{aligned} \frac{dy}{dx} &= \frac{1 + y^{2}}{\tan^{-1} y - x} \end{aligned} \]

This equation is not readily solvable. However, we can instead try to get \(x\) a function of \(y\) by inverting the above equation

\[ \begin{aligned} \frac{dx}{dy} &= \frac{\tan^{-1} y - x}{1 + y^{2}}\newline &= \frac{\tan^{-1} y}{1 + y^{2}} - \frac{1}{1 + y^{2}}x\newline \implies \frac{dx}{dy} + \frac{1}{1 + y^{2}}x &= \frac{\tan^{-1} y}{1 + y^{2}} \end{aligned} \]

Which is a linear equation. The integrating factor is

\[ \begin{aligned} IF &= e^{\int \frac{1}{1 + y^{2}}dy} = e^{\tan^{-1} y}\newline \implies e^{\tan^{-1} y}\frac{dx}{dy} + \frac{e^{\tan^{-1} y}}{1 + y^{2}}x &= \frac{e^{\tan^{-1} y}\tan^{-1} y}{1 + y^{2}}\newline xe^{\tan^{-1} y} &= \int \frac{e^{\tan^{-1} y}\tan^{-1} y}{1 + y^{2}} dy\newline \end{aligned} \]

Substituiting \(\tan^{-1} y = v\)

\[ \begin{aligned} \frac{1}{1 + y^{2}}dy &= dv\newline \implies \int \frac{e^{\tan^{-1} y}\tan^{-1} y}{1 + y^{2}} &= \int te^{t}dt\newline &= te^{t} - e^{t} + c\newline &= e^{\tan^{-1} y}\tan^{-1} y - e^{\tan^{-1} y} + c \end{aligned} \]

Thus,

\[ \begin{aligned} xe^{\tan^{-1} y} &= e^{\tan^{-1} y}\tan^{-1} y - e^{\tan^{-1} y} + c\newline \implies x &= \tan^{-1} y - 1 + ce^{-\tan^{-1} y} \end{aligned} \]