Answer to Question #220266 in Differential Equations for Sarak

Question

(1+yz)dx + z(z-x)dy - ( 1 + xy ) dz = 0 verify that the pfaffian differential equation are intergrable and find corresponding integral

Answer 1

This is a Pfaffian differential equation in three variables and we must verify  its integrabilty and determine its primitive.

$$(1+yz)dx+z(z−x)dy−(1+xy)dz=0$$

The necessary and sufficient condition for integrability is

$$X⋅curlX=0$$

$$X=(1+yz,z^2-xz,−1−xy),$$

so that

$$\nabla \times X=\begin{vmatrix} i & j & k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ 1+yz & z^2-xz & -1-xy \\ \end{vmatrix}$$

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

$$=-2zi+2yj-2zk$$

$$=(1+yz,z^2-xz,−1−xy)\cdot(-2z, 2y, -2z)$$

$$=-2z-2yz^2+2yz^2-2xyz+2z+2xyz=0$$

Thus, the given equation is integrable.

Solving by Inspection

$$(1+yz)dx+z(z−x)dy−(1+xy)dz=0$$

$$(1+yz)dx-(1+yz)dz+(1+yz)dz$$

$$−(1+xy)dz+z(z−x)dy=0$$

$$-(1+yz)d(z-x)+y(z-x)dz+z(z-x)dy=0$$

$$-(1+yz)d(z-x)+(z-x)d(1+yz)=0$$

$$\dfrac{d(1+yz)}{1+yz}-\dfrac{d(z-x)}{z-x}=0$$

Integrating both sides, we have

$$\ln(1+yz)-\ln(z-x)=\ln C$$

$$1+yz=C(z-x)$$

$$z=\dfrac{1+Cx}{C-y}$$

is a solution to the Pfaffian differential equation.