We will first write this equation as

\(x \arctan (\frac{y}{x}) \cdot \frac{dy}{dx} = y \arctan (\frac{y}{x}) - x\) (1)

Notice that \(\arctan (\frac{y}{x}) = 0\)

means that \(\frac{y}{x} = 0,\ so\ y = 0.\) However, this is

not a solution of the given equation.

Thus, \(\arctan (\frac{y}{x}) \neq 0,\) so we can divide the equation (1)

by \(x \arctan (\frac{y}{x})\)

\(\frac{dy}{dx}=\frac{y}{x}-\frac{1}{\arctan(\frac{y}{x})}\) (2)

Make the substitution \(u = \frac{y}{x}\ or\ y = ux.\) Then

\(\frac{dy}{dx} = \frac{d}{dx} (ux) = x \frac{dy}{dx} + u\) (3)

(use the Chain Rule). Furthermore, (2) becomes

\(\frac{dy}{dx} = u - \frac{1}{\arctan(u)}\) (4)

Combining (3) and (4), we get \(x \frac{du}{dx} + u = u - \frac{1}{\arctan(u)} \Rightarrow x \frac{du}{dx} = - \frac{1}{\arctan(u)}\)

Notice that this is a separable equation! It can be written as

\(\arctan(u) du = - \frac{1}{x} dx\)

Integrate both sides:

\(\int \arctan(u) du = - \int \frac{dx}{d}\) (5)

Clearly,

\(- \int \frac{dx}{d} = - \ln |x| + C_{2},\)

where \(C_{2}\) is some constant.

For the other integral, we use the integration by parts:

\(\int \arctan(u) du = \{(t = \arctan(u) dv = du), (dt = \frac{du}{1+u^{2}} v = u)\}\)

\(= u \arctan(u) - \int \frac{udu}{1+u^{2}} du\)

\(= u \arctan(u) - \frac{1}{2} \ln |1 + u^{2}| + C_{1}\)

\(= u \arctan(u) - \ln \sqrt{1+u^{2}} + C_{1}\)

where \(C_{1}\) is some constant.

(In last equality we need used that \(1 + u^{2} > 0,\ so\ |1 + u^{2}| = 1 + u^{2},\)

and the property of the logarithmic function: \(x \ln y = \ln y^{x}.)\)

Finnaly. (5) becomes

\(= u \arctan(u) - \ln \sqrt{1+u^{2}} = - \ln |x| + C,\)

where \(C = C_{2} - C_{1}\)