Question

Verify that the equation 2(y+z)dx - (x+z)dy+ (2y-x+z)dz = 0 is integrable and find its primitive .

Answer 1

We have

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

General form of the Pfafffian equation is

$$Pdx+Qdy+Rdz=0$$

The integrability condition for the Pfaffian equation is

$$(curlF, F)=0$$

where $F=(P, Q, R),$ or

$$P({\eth Q \over \eth z}-{\eth R \over \eth y})+Q({\eth R \over \eth x}-{\eth P \over \eth z})+R({\eth P \over \eth y}-{\eth Q \over \eth x})=0$$

Verify

$$P=2(y+z), Q=-(x+z), R=(2y-x+z)$$ $$2(y+z)(-1-2)-(x+z)(-1-2)+(2y-x+z)(2+1)=$$ $$=-6y-6z+3x+3z+6y-3x+3z=0$$

The integrability condition for this equation hold.

If the Pfaffian equation is multiplied by a certain function $\mu(x, y,z)$ then one can obtain in the

left-hand side the total differential.

If we treat $z$ as a constant, then the given equation reduces to

$$2(y+z)dx-(x+z)dy=0$$ $${2 \over x+z}dx-{1 \over y+z}dy=0$$

which has a solution

$$U(x, y, z)={(x+z)^2 \over y+z}=C_1$$

The integrating factor $\mu(x,y,z)$ is given by

$$\mu={1 \over P}\cdot {\eth U \over \eth x}={1 \over 2(y+z)}\cdot {2(x+z) \over y+z}={x+z \over (y+z)^2}$$

We have

$$K=\mu R-{\eth U \over \eth z}=$$ $$={x+z \over (y+z)^2}(2y-x+z)-{2(x+z)(y+z)-(x+z)^2 \over (y+z)^2}=$$ $$={x+z \over (y+z)^2}(2y-x+z-2y-2z+x+z)=0$$

Thus we have the equation

$$dU=0$$

which has a solution $U=c$

Thus the solution of the given equation

$$(x+z)^2=c(y+z)$$

where $c$ is a constant.