$2(2x^2 + y^2) \, dx - xy \, dy = 0$

Let

$y = vx$

$dy = v \, dx + x \, dv$

$2(2x^2 + v^2x^2) \, dx - vx^2(v \, dx + x \, dv) = 0$

$4x^2 \, dx + 2v^2x^2 \, dx - v^2x^2 \, dx - vx^3 \, dv = 0$

$4x^2 \, dx + v^2x^2 \, dx - vx^3 \, dv = 0$

$x^2(4 + v^2) \, dx - vx^3 \, dv = 0$

$\dfrac{x^2(4 + v^2) \, dx}{x^3(4 + v^2)} - \dfrac{vx^3 \, dv}{x^3(4 + v^2)} = 0$

$\dfrac{dx}{x} - \dfrac{v \, dv}{4 + v^2} = 0$

$\displaystyle \int \dfrac{dx}{x} - \dfrac{1}{2} \int \dfrac{2v \, dv}{4 + v^2} = 0$

$\ln x - \frac{1}{2} \ln (4 + v^2) = \ln c$

$2\ln x - \ln (4 + v^2) = 2\ln c$

$\ln x^2 - \ln (4 + v^2) = \ln c^2$

$\ln x^2 = \ln c^2 + \ln (4 + v^2)$

$\ln x^2 = \ln c^2(4 + v^2)$

$x^2 = c^2(4 + v^2)$

From

$y = vx$

$v = \dfrac{y}{x}$

Thus,

$x^2 = c^2\left( 4 + \dfrac{y^2}{x^2} \right)$

$x^2 = c^2\left( \dfrac{4x^2 + y^2}{x^2} \right)$

$x^4 = c^2(4x^2 + y^2)$ *answer*