Question

Find the general integral of
(x^3-3xy^2)p+(y^3+3x^2y)q=2(x^2+y^2)z

Answer 1

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

$\frac{dy}{dx}=\frac{y^3+3x^2y}{x^3-3xy^2}$

This is a homogeneous differential equation.

Let $y/x=v, y=xv, dy/dx=v+x(dv/dx)$

Then:

$v+x\frac{dv}{dx}=\frac{(xv)^3+3x^3v}{x^3-3x^3v^2}=\frac{v^3+3v}{1-3v^2}$

$x\frac{dv}{dx}=\frac{v^3+3v}{1-3v^2}-v=\frac{4v^3+2v}{1-3v^2}$

$\frac{1}{2}\intop\frac{1-3v^2}{v(2v^2+1)}dv=\intop \frac{dx}{x}$

$\frac{1-3v^2}{v(2v^2+1)}=\frac{A}{v}+\frac{Bv+C}{2v^2+1}$

$A(2v^2+1)+v(Bv+C)=1-3v^2$

$2A+B=-3$

$A=1$

$C=0$

$B=-3-2A=-5$

$\frac{1-3v^2}{v(2v^2+1)}=\frac{1}{v}-\frac{5v}{2v^2+1}$

$\intop\frac{v}{2v^2+1}dv=\frac{1}{2}\intop\frac{d(v^2)}{2v^2+1}dv=\frac{ln(2v^2+1)}{4}$

$\frac{lnv}{2}-\frac{5ln(2v^2+1)}{8}=lnx+lnc_1$

$ln(\frac{v^{1/2}}{(2v^2+1)^{5/8}})=ln(c_1x)$

$\frac{(y/x)^{1/2}}{x(2(y/x)^2+1)^{5/8}}=c_1$

$\frac{dx/x+dy/y+dz/z}{6x^2}=-\frac{3(dx/x-dy/y+dz/z)}{6y^2}=\frac{dz}{2z(x^2+y^2)}$

$\frac{dx/x+dy/y+dz/z-3(dx/x-dy/y+dz/z)}{6x^2+6y^2}=\frac{3dz/z}{6(x^2+y^2)}$

$dx/x+dy/y+dz/z-3(dx/x-dy/y+dz/z)=3dz/z$

$4dy/y-2dx/x=5dz/z$

$4lny-2lnx=5lnz+lnc_2$

$ln(y^4/x^2)=ln(c_2z^5)$

$\frac{y^4}{x^2z^5}=c_2$

The general integral:

$\phi(\frac{(y/x)^{1/2}}{x(2(y/x)^2+1)^{5/8}}, \frac{y^4}{x^2z^5})=0$