Solve the homogeneous equation: y^3 dx+2(x^3-xy^2 )dy=0.
Divide through by y^3:
dx+2(x^3/y^3-x/y)dy=0
This is homogeneous in x/y, so let u=x/y or uy=x, and udy+ydu=dx. Therefore substitute for dx:
udy+ydu+2(u^3-u)dy=0=ydu+(2u^3-u)dy=0.
ydu=(u-2u^3)dy; du/(u-2u^3)=dy/y; so ln|y|=∫du/(u-2u^3).
u-2u^3 factorises: u(1-2u^2)=u(1+u√2)(1-u√2) and break the integral into partial fractions:
A/u+B/(1+u√2)+C/(1-u√2)=(A(1-2u^2)+B(u-u^2√2)+C(u+u^2√2))/(u-2u^3)=1/(u-2u^3).
The only constant is A so A=1.
Equating coefficients:
u: B+C=0, so C=-B
u^2: -2A-B√2+C√2=0=-2A-B√2-B√2=0=-2-2B√2; 1+B√2=0, so B=-√2/2, and C=√2/2.
∫du/(u-2u^3)=∫du/u-(√2/2)∫du/(1+u√2)+(√2/2)∫du/(1-u√2)=
ln|u|-(1/2)ln|1+u√2|-(1/2)ln|1-u√2|=ln|(u/√(1-2u^2)|. [(1/2)ln|1+u√2|=ln(√(1+u√2)).]
We can include the constant of integration by writing ln|(Au/√(1-2u^2)|, where A is a constant, because the constant c can be replaced by A=ln(c).
So we have ln|y|=ln|(Au/√(1-2u^2)|; y=Au/√(1-2u^2), because the log arguments must be equal.
Finally we replace u by x/y:
y=A(x/y)/√(1-2x^2/y^2)=Ax/√(y^2-2x^2) (implicit equation).
This can be written as the quartic: y^4-2x^2y^2-A^2x^2=0.
[y^2=A^2x^2/(y^2-2x^2), y^4-2y^2x^2-A^2x^2=0=y^4-2y^2x^2+x^4-x^4-A^2x^2.
(y^2-x^2)^2=x^4+A^2x^2; y^2-x^2=±x√(x^2+A^2); y=x^2±x√(x^2+A^2); but it may be more appropriate to write x in terms of y: x^2=y^4/(2y^2+A^2) or x=±y^2/√(2y^2+A^2).]
If we want y in terms of x, this could be expressed as the solution of a quadratic:
y^2=x^2±x√(x^2+A^2), or graphed as a double parabola as shown in the graph below:
y^2-x^2±x√(x^2+A^2)=0 or y^4-2y^2x^2-A^2x^2=0.
The graph appears to be something like (A has been given the arbitrary value 10 for the purpose of illustration):
The various colours arise from ±. The graph is basically two horizontal parabolas touching at their vertices.
CHECK THE SOLUTION USING IMPLICIT DIFFERENTIATION
We arrived at the solution y=Ax/√(y^2-2x^2).
We can differentiate by using the identity d(u/v)/dx=(vu'-uv')/v^2, where u=Ax and v=√(y^2-2x^2).
So u'=A, v'=(2yy'-4x)/(2√(y^2-2x^2))=(yy'-2x)/√(y^2-2x^2).
Differentiating we get:
y'=(vu'-uv')/v^2=[A√(y^2-2x^2)-Ax(yy'-2x)/√(y^2-2x^2)]/(y^2-2x^2).
y'(y^2-2x^2)=[A(y^2-2x^2)-Ax(yy'-2x)]/√(y^2-2x^2)=A(y^2-2x^2-xyy'+2x^2)/√(y^2-2x^2)=A(y^2-xyy')/√(y^2-2x^2).
But y=Ax/√(y^2-2x^2), so A/√(y^2-2x^2)=y/x.
Therefore, y'(y^2-2x^2)=y(y^2-xyy')/x.
We need to find y' in terms of the other variables.
y'(xy^2-2x^3)=y^3-xy^2y'; y'(xy^2-2x^3+xy^2)=y^3=2y'(xy^2-x^3).
Rewriting y as dy/dx: y^3=2dy/dx(xy^2-x^3) or y^3dx+2(x^3-xy^2)dy=0, the original DE.
So the general solution, as an implicit equation, y=Ax/√(y^2-2x^2) is correct.