$\frac{dy}{dx} = 2x(x + y)^3 - x(x + y) - 1$

$\implies \quad 1 + \frac{dy}{dx} = 2x(x + y)^3 - x(x + y) \qquad \dots \text{(i)}$

$\text{Let } x + y = z$

$\implies \quad 1 + \frac{dy}{dx} = \frac{dz}{dx}$

$\text{From (i), } \frac{dz}{dx} = 2xz^3 - xz$

$\implies \quad z^{-3} \frac{dz}{dx} + xz^{-2} = 2x \qquad \dots \text{(ii)}$

$\text{Now put } z^{-2} = v \implies -2z^{-3} \frac{dz}{dx} = \frac{dv}{dx}$

$\text{From (ii) } \implies \quad -\frac{1}{2} \frac{dv}{dx} + vx = 2x \quad \implies \quad \frac{dv}{dx} - 2vx = -4x \to \text{LDE}$

$\text{I.F.} = e^{-\int 2x \, dx} = e^{-x^2}$

$\text{Solution of LDE}$

$v(e^{-x^2}) = \int -4x \cdot e^{-x^2} \, dx \; ; \quad v e^{-x^2} = 2e^{-x^2} + c$

$\implies \quad \frac{1}{(x + y)^2} e^{-x^2} = 2e^{-x^2} + c$

$\because \quad y(0) = 1 \quad \implies \quad 1 = 2 + c \quad \implies \quad c = -1$

$\therefore \quad \frac{1}{(x + y)^2} e^{-x^2} = 2e^{-x^2} - 1, \quad \frac{1}{(x + y)^2} = 2 - e^{x^2} \quad \text{or} \quad (x + y)^2 = \frac{1}{2 - e^{x^2}}$

$\text{Put } x = \frac{1}{\sqrt{2}} \quad \implies \quad \left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^2 = \frac{1}{2 - \sqrt{e}}$

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