Answer to Question #194095 in Differential Equations for jack

Question

Solve : 2(2y^2+yz -z^2)dx+x(4y+z)dy+x(y-2z)dz= 0

Answer 1

$2(2y^2+yz -z^2)dx+x(4y+z)dy+x(y-2z)dz= 0$

$2y^2+yz -z^2=(2y-z)(y+z)$

Put $2y-z=u$ , $y+z=v$ . Then $u+v=3y$ and $2v-u=3z$

$4y+z = (4(u+v)+(2v-u))/3 = u+2v$

$y-2z = ((u+v)-2(2v-u))/3 = u-v$

$2(2y^2+yz -z^2)dx+x(4y+z)dy+x(y-2z)dz =$

$= 2uvdx+x(u+2v)d(u+v)/3 + x(2v-u)d(u-v)/3 =$

$= 2uvdx+x((u+2v) - (u-v))du/3 + x((u+2v) +2 (u-v))dv/3 =$

$= 2uvdx+xvdu + xudv = 0$

Special solutions: (i) $x=0$ , (ii) $u=0$ , (iii) $v=0$

Assuming $x\ne 0$ , $u\ne 0$ and $v\ne 0$ , we may write then

$0 = 2uvdx+xvdu + xudv=xuv(2\frac{dx}{x} +\frac{du}{u}+\frac{dv}{v})=xuv\cdot d\log|x^2uv|$

$x^2uv=C$ (the general solution)

The special solutions (i) - (iii) satisfy the equation of the general solution for C=0.

Answer. $x^2(2y-z)(y+z)=C$