I think this is supposed to be:
2x³ydx+(x⁴+y⁴)dy=0, because it would not make sense otherwise.
If we make the substitution y=vx then dy/dx=v+xdv/dx, where v is a function of x.
We then have:
2vx⁴dx+(x⁴+v⁴x⁴)dy=0 and assuming at this point that x≠0, we can divide through by x⁴:
2vdx+(1+v⁴)dy=0⇒(1+v⁴)dy=-2vdx⇒dy/dx=-2v/(1+v⁴).
Now replace dy/dx by v+xdv/dx:
v+xdv/dx=-2v/(1+v⁴). We can separate variables x and v:
v+2v/(1+v⁴)=-xdv/dx⇒(v+v⁵+2v)/(1+v⁴)=-xdv/dx.
This can be written:
[(1+v⁴)/(3v+v⁵)]dv=-dx/x ready for integration.
(1+v⁴)/(3v+v⁵) is (1+v⁴)/(v(3+v⁴)) and split into partial fractions:
A/v+Bv³/(3+v⁴) where A and B are constants to be found:
A(3+v⁴)+Bv⁴≡1+v⁴, so equating coefficients: A+B=1 (v⁴), 3A=1 (constant), so A=⅓ and B=⅔.
We need to solve:
∫(⅓/v+⅔v³/(3+v⁴))dv=-∫dx/x.
The left hand integral can be split:
⅓∫dv/v+⅔∫(v³/(3+v⁴))dv=-∫dx/x.
If we let u=3+v⁴, du=4v³dv, so v³dv=du/4.
So we have:
⅓ln|v|+⅔∫(1/u)(du/4)=-ln|x|+c where c is constant of integration.
⅓ln|v|+⅙ln|u|=ln|a/x| where c=ln(a), that is, a=e^c, a constant.
⅓ln|v|+⅙ln(3+v⁴)=ln|a/x|.
ln(v²(3+v⁴))=ln(A/x⁶), where A=a⁶,
v²(3+v⁴)=A/x⁶
But v=y/x, so (y/x)²(3+(y/x)⁴)=A/x⁶,
y²(3x⁴+y⁴)=A.
