From the question it is given that, \( (\dfrac{dy}{dx}) =\dfrac{ (y – x)}{(y + x)}\)

The given differential equation is a homogeneous equation,

Let us assume, y = vx and \( (\dfrac{dy}{dx}) = v + x (\dfrac{dv}{dx})\)

\( v + x (\dfrac{dv}{dx})=\dfrac{(vx – x)}{(vx + x)}\)

Then, \( v + x (\dfrac{dv}{dx})=\dfrac{(v – 1)}{(v + 1)}\)

\( x (\dfrac{dv}{dx}) = \dfrac{(v – 1)}{(v + 1)} – v\)

\( x (\dfrac{dv}{dx}) = \dfrac{(v – 1 – v^2 – v)}{(v + 1)}\)

On dividing we get,

\( x (\dfrac{dv}{dx}) =\dfrac{ – (1 + v^2)}{(v + 1)}\)

Now, taking like variables on same side,

\( \dfrac{(v + 1)}{(v^2 + 1)} = -\dfrac{dx}{x}\)

Integrating on both side we get,

\( \int \dfrac{(v + 1)}{(v^2 + 1)}dv = -\int \dfrac{dx}{x}\)

Then,

\( log \dfrac{(y^2 + x^2)}{x^2}+ 2 tan^{-1} (\dfrac{y}{x}) = 2 log (\dfrac{c}{x})\)

log [x^{2} + y^{2}] – 2 log x + 2 tan^{-1} (\( \dfrac{y}{x}\)) = 2 log (\( \dfrac{c}{x}\))

log (x^{2} + y^{2}) + 2 tan^{-1} (\( \dfrac{y}{x}\)) = 2 log c

log (x^{2} + y^{2}) + 2 tan^{-1} (\( \dfrac{y}{x}\)) = k

Find the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)

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