Very hard differential equation. Help please please :(

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

The answer will be

x^4 = c^2(4x^2+y^2)

I used homogeneous and that y = vx and dy = vdx + x dv
but it didnt work out with me.. can u please use the substitution method
Thats what i did:
2(2x^2 + v^2 x^2) dxd - x (vx)(vdx + xdv ) = 0
I simplified it till i reached
1/x dx + v/4+v^2 dv = 0
but when I integrate it i got a different answer, can you please help me out!
I deeply appreciate it
Thank you so much for your time

Make a substitution to form a separable differential equation:
2(2x² + y²) dx - xy dy = 0
2(2x² + y²) - xy(dy / dx) = 0
xy(dy / dx) = 2(2x² + y²)
dy / dx = 2(2x / y + y / x)
dy / dx = 4x / y + 2y / x
dy / dx = 4 / (y / x) + 2y / x
Let u = y / x,
y = ux
dy / dx = x(du / dx) + u
x(du / dx) + u = 4 / u + 2u

Solve this differential equation by separating the variables then integrating:
x(du / dx) = 4 / u + u
x(du / dx) = (4 + u²) / u
u / (4 + u²) du = dx / x
∫ 2u / (4 + u²) du / 2 = ∫ 1 / x dx
ln(4 + u²) / 2 = ln|x| + C
ln(4 + u²) / 2 = C + ln|x|
ln(4 + u²) = C + 2ln|x|
4 + u² = ℮^(C + 2ln|x|)
4 + u² = ℮^(C + lnx²)
4 + u² = ℮ᶜx²
4 + u² = Cx²
u² = Cx² - 4
u = ±√(Cx² - 4)

Find the general solution by substituting back for the previous variables:
Since u = y / x,
y / x = ±√(Cx² - 4)
y = ±x√(Cx² - 4)

2(2x^2+y^2) dx - xy dy = 0
4x^2 dx + 2y^2 dx - xy dy =0

divide both sides by dy
4x^2 + 2y^2 - xy dy/dx = 0

Rewrite the equation
- x dy/dx y + 2y^2 = -4x^2

divide both sides by -x/2
2 dy/dx y - 4y^2 /x = 8x ------------(1)

Let v = y^2
dv/dx = 2y dy/dx

Equation (1) becomes
dv/dx - 4v /x = 8x -------------(2)

This is a linear equation of first order of the form
dy/dx + v p(x) = q(x)

p(x) - -4/x
q(x)=8x

Find the integrating factor e^∫p(x) dx = e^∫ -4/x dx = -e^[4 ln(x)] = e^ln(x^-4) = 1/x^4