Question

(y^2 +yz+ z^2)dx +(z^2 +zx+ x^2)dy + (x^2 +xy+ y^2)dz = 0

Answer 1

Verify the Pfaffian Differential equation

$(y^2+yz+z^2)dx+(z^2+zx+x^2)dy+(x^2 +xy+y^2)dz=0$

is integrable and find its primitive.

The necessary and sufficient condition for iintegrability is

$$X\cdot curlX=0,$$

$X=(y^2+yz+z^2,z^2+zx+x^2,x^2 +xy+y^2)$

So that

$$\nabla \times X=\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ y^2+yz+z^2 & z^2+zx+x^2 & x^2 +xy+y^2 \end{vmatrix}$$

$$=\mathbf i(x+2y-2z-x)- \mathbf j(2x+y-y-2z)+$$ $$+\mathbf k(z+2x-2y-z)$$ $$=2(y-z)\mathbf i+2(z-x) \mathbf j+2(x-y) \mathbf k$$

$$X\cdot curlX=2(y^2+yz+z^2)(y-z)+$$ $$+2(z^2+zx+x^2)(z-x)+2(x^2 +xy+y^2)(x-y)=$$ $$=2(y^3-y^2z+y^2z-yz^2+yz^2-z^3)+$$ $$+2(z^3-xz^2+xz^2-x^2z+x^2z-x^3)+$$ $$=2(x^3-x^2y+x^2y-xy^2+xy^2-y^3)=0$$

Thus the given equation is integrable.

$$P=y^2+yz+z^2,$$ $$Q=z^2+zx+x^2,$$ $$R=x^2 +xy+y^2.$$

The auxiliary equations are

$${dx \over {\partial Q\over \partial z}-{\partial R \over \partial y}}={dy \over {\partial R\over \partial x}-{\partial P \over \partial z}}={dz \over {\partial P\over \partial y}-{\partial Q \over \partial x}}$$

$${dx \over 2(z-y)}={dy \over 2(x-z)}={dz \over 2(y-x)}$$

$${dx \over z-y}={dy \over x-z}={dz \over y-x}$$

$$dx+dy+dz=0=>x+y+z=c_1=u$$

$$(y+z)dx+(x+z)dy+(x+y)dz=0=>xy+yz+zx=c_2=v$$

Formulate the equation $Adu+Bdv=0$ and compare with the given equation.

$Adx+Ady+Adz+Bxdy+Bydx+Bzdy+Bydz+Bzdx+Bxdz=0$

$$A+By+Bz=y^2+yz+z^2$$ $$A+Bx+Bz=z^2+xz+x^2$$ $$A+By+Bx=x^2+xy+y^2$$

$$B=x+y+z$$ $$A=-xy-yz-xz$$

Then

$$A=-v$$ $$B=u$$

$$Adu+Bdv=0$$ $$-vdu+udv=0$$

$${dv \over v}={du \over u}$$

$$\ln v=\ln u+\ln c$$ $$v=cu$$

The required solution is

$$xy+yz+zx=c(x+y+z)$$