Question

Find the integral of the Pfaffian differential equation
(1+yz)dx+x(z-x)dy-(1+xy)dz=0

Answer 1

The necessary and sufficient conditionfor iintegrability is

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

$-(1+xy)(z-(z-2x))+(1+yz)(x+x)+x(z-x)(-y-y)=$

$=-2x-2x^2y+2x+2xyz-2xyz+2x^2y=0$

Thus, the given equation is integrable.

$(1+yx)dz−(1+yx)dx−x(z−x)dy−y(z−x)dx=0$

$(1+yx)d(z−x)−(xdy+ydx)(z−x)=0$

$(1+yx)d(z−x)−d(1+xy)(z−x)=0$

$\frac{d(z-x)}{z-x}-\frac{d(1+xy)}{1+xy}=0$

$ln(z−x)−ln(1+xy)=lnc_1 ​$

$\frac{z-x}{1+xy}=c_1$