Question

Find the integrals of the Pfaffian differential equation
z(z-y)dx+z(x+z)dy+x(x+y)dz=0

Answer 1

Given differential equation is

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

Let $\vec{X} = z(z-y)\hat{i} +z(x+z)\hat{j}+x(x+y)\hat{k}$

To check whether it is integral ,$\vec{X} .(\nabla\times{\vec{X}}) = 0$

Now, $\nabla\times{\vec{X}} = \begin{vmatrix} \hat{i} & \hat{j} &\hat{k}\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z}\\ z(z-y) & z(x+z) & x(x+y) \end{vmatrix}$

$\nabla\times{\vec{X}} = -2z\hat{i} +2(x+y-z)\hat{j} + (z+1)\hat{k}$

Now, $\vec{X} .(\nabla\times{\vec{X}}) =[ z(z-y)\hat{i} +z(x+z)\hat{j}+x(x+y)\hat{k} ] .[-2z\hat{i} +2(x+y-z)\hat{j} + (z+1)\hat{k}] = 0$

Hence it is integrable.

Let $y = ux \implies dy = xdu + udx$

and $z = vx \implies dz = vdx+xdv$

Then partial differential equation will be reduced to the form

$vx(vx-ux)dx + vx(x+vx)(udx+xdu) + x(x+ux)(vdx+xdv) = 0$

It can be further reduced to

$\frac{dx}{x} + \frac{du}{1}+\frac{dv}{v}-\frac{du}{(u+2v+vu)}-\frac{dv}{(u+2v+uv)} = 0$

Integrating it, we get

$lnx+u+lnv-\frac{ln(2v+u(1+v))}{v+1} -\frac{ln(u+v(2+u))}{2+u} = C$

putting values of u and v,

$lnz+\frac{y}{x}-\frac{ln(2\frac{z}{x}+\frac{y}{x}(1+\frac{z}{x}))}{v+1} -\frac{ln(\frac{y}{x}+\frac{z}{x}(2+u))}{2+\frac{y}{x}} = C= C(x+y+z)$