2(2y2+yz−z2)dx+x(4y+z)dy+x(y−2z)dz=0
2y2+yz−z2=(2y−z)(y+z)
Put 2y−z=u , y+z=v . Then u+v=3y and 2v−u=3z
4y+z=(4(u+v)+(2v−u))/3=u+2v
y−2z=((u+v)−2(2v−u))/3=u−v
2(2y2+yz−z2)dx+x(4y+z)dy+x(y−2z)dz=
=2uvdx+x(u+2v)d(u+v)/3+x(2v−u)d(u−v)/3=
=2uvdx+x((u+2v)−(u−v))du/3+x((u+2v)+2(u−v))dv/3=
=2uvdx+xvdu+xudv=0
Special solutions: (i) x=0 , (ii) u=0 , (iii) v=0
Assuming x=0 , u=0 and v=0 , we may write then
0=2uvdx+xvdu+xudv=xuv(2xdx+udu+vdv)=xuv⋅dlog∣x2uv∣
x2uv=C (the general solution)
The special solutions (i) - (iii) satisfy the equation of the general solution for C=0.
Answer. x2(2y−z)(y+z)=C
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